Encuentra la forma cerrada de la integral cuádruple

Definir:

$$\begin{eqnarray} I(a,b,c,d):=\int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta \quad (i) \end{eqnarray}$$ Luego también define:

$$\begin{eqnarray} {\mathfrak F}^{(A,B)}_{a,b} &:=& \int\limits_A^B \frac{\log(z+a)}{z+b} dz\\ &=& F[B,a,b] - F[A,a,b] + 1_{t^* \in (0,1)} \left( -F[A+(t^*+\epsilon)(B-A),a,b] + F[A+(t^*-\epsilon)(B-A),a,b] \right) \quad (ii) \end{eqnarray}$$ dónde $$\begin{eqnarray} t^*:=-\frac{Im[(A+b)(b^*-a^*)]}{Im[(B-A)(b^*-a^*)]} \end{eqnarray}$$ y $$\begin{equation} F[z,a,b] := \log(z+a) \log\left( \frac{z+b}{b-a}\right) + Li_2\left( \frac{z+a}{a-b}\right) \end{equation}$$ para$a$,$b$,$A$,$B$siendo complejo.

Entonces nosotros tenemos:

$$\begin{eqnarray} I(a,b,c,d)&=& \frac{1}{\sqrt{a}} \int\limits_0^{\sqrt{\frac{d}{a+b+c}}} \frac{u \tan ^{-1}(u)}{\left(d-c u^2\right) \sqrt{d-u^2 (b+c)}} du\\ &=& \frac{\sqrt{d}}{\sqrt{a} (b+c)}\int\limits_0^{\sin ^{-1}\left(\sqrt{\frac{b+c}{a+b+c}}\right)} \frac{\sin (\phi ) \tan ^{-1}\left(\sin (\phi ) \sqrt{\frac{d}{b+c}}\right)}{d-\frac{c d \sin ^2(\phi )}{b+c}} d\phi\\ &=&-\frac{2 i}{\sqrt{a} \sqrt{d}} \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{t}{b \left(t^2+1\right)^2+c \left(t^2-1\right)^2} \log \left(\frac{2 i t \sqrt{\frac{d}{b+c}}+t^2+1}{-2 i t \sqrt{\frac{d}{b+c}}+t^2+1}\right) dt\\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{\log \left(i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}}+t\right)}{t-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} dt \\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } % {\mathfrak F}^{(0,\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1})}_{i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}},-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} \end{eqnarray}$$

En la línea superior sustituimos$u=\sin(\theta) \sqrt{d/(b+c \sin(\theta)^2)}$. En la segunda línea sustituimos$u = \sqrt{d/(c+b)} \sin(\phi)$. En la tercera línea sustituimos$t=\tan(\phi/2)$. En la cuarta línea usamos descomposición en fracciones parciales y propiedades del logaritmo. Finalmente en la quinta línea usamos la antiderivada definida en$(ii)$.

Clear[F]; Clear[FF];
F[z_, a_, b_] := 
  Log[a + z] Log[(b + z)/(-a + b)] + PolyLog[2, (a + z)/(a - b)];
FF[A_, B_, a_, b_] := 
  Module[{result, ts, zs, zsp, zsm, eps = 10^(-15)},
   (*This is Integrate[Log[z+a]/(z+b),{z,A,B}] where all a,b,A, 
   and B are complex. *)
   result = F[B, a, b] - F[A, a, b];


   ts = - (Im[(A + b) (Conjugate[b] - Conjugate[a])]/
     Im[(B - A) (Conjugate[b] - Conjugate[a])]);
   If[0 <= ts <= 1,
    zsp = A + (ts + eps) (B - A);
    zsm = A + (ts - eps) (B - A);
    result += -F[zsp, a, b] + F[zsm, a, b];
    ];

   result
   ];

{a, b, c, d} = RandomReal[{0, 3}, 4, WorkingPrecision -> 50];
NIntegrate[
 Exp[-a/2 x^2 - b/2 y^2 - c/2 z^2 - d/2 w^2], {x, 0, Infinity}, {y, 0,
   x}, {z, 0, y}, {w, 0, z}]
NIntegrate[
 Sin[th]/Sqrt[a b d (b + c Sin[th]^2)] ArcTan[
   Sin[th] Sqrt[d/(b + c Sin[th]^2)]], {th, 0, ArcTan[Sqrt[b/a]]}]
  1/Sqrt[a ] NIntegrate[
  u ArcTan[u] 1/((d - c u^2) Sqrt[d - (c + b) u^2]), {u, 0, Sqrt[ 
   d/ (a + b + c)]}]
 Sqrt[d]/(Sqrt[a ] (b + c))
  NIntegrate[
  Sin[phi] ArcTan[
    Sqrt[d/(c + b)] Sin[phi]] 1/(d - c (d/(c + b) Sin[phi]^2)) , {phi,
    0, ArcSin[ Sqrt[( c + b)/ (a + b + c)]]}]
- I 2/(Sqrt[a ] Sqrt[d])
  NIntegrate[
   t /(c (-1 + t^2)^2 + b (1 + t^2)^2) Log[(
    1 + t^2 + 2 I Sqrt[d/(b + c)] t)/(
    1 + t^2 - 2 I Sqrt[d/(b + c)] t)], {t, 0, Sqrt[(b + c)/(
   a + b + c)]/(1 + Sqrt[a/(a + b + c)])}]
- I 2/(Sqrt[a ] Sqrt[d])
  NIntegrate[
   t /(c (-1 + t^2)^2 + 
     b (1 + t^2)^2) Log[((1/
        2 (2 I Sqrt[d/(b + c)] - Sqrt[-4 - (4 d)/(b + c)]) + 
       t) (1/2 (2 I Sqrt[d/(b + c)] + Sqrt[-4 - (4 d)/(b + c)]) + 
       t))/((1/2 (-2 I Sqrt[d/(b + c)] - Sqrt[-4 - (4 d)/(b + c)]) + 
       t) (1/2 (-2 I Sqrt[d/(b + c)] + Sqrt[-4 - (4 d)/(b + c)]) + 
       t))], {t, 0, Sqrt[(b + c)/(a + b + c)]/(
   1 + Sqrt[a/(a + b + c)])}]
 1/Sqrt[a b c d]  1/4 NIntegrate[
  Sum[(-1)^(Floor[(eta - 1)/2]) (-1)^
    Floor[(xi - 1)/2] Log[
     t + (-1)^Floor[(xi - 1)/2] I Sqrt[d/(b + c)] + (-1)^(xi - 1)
        I Sqrt[( b + c + d)/(b + c)]]/(
    t - (-1)^(1 + eta + 
        Floor[(eta - 1)/2]) I Exp[(-1)^(Floor[(eta - 1)/2]) I ArcTan[
         Sqrt[c]/Sqrt[b]]]), {xi, 1, 4}, {eta, 1, 4}], {t, 0, Sqrt[(
   b + c)/(a + b + c)]/(1 + Sqrt[a/(a + b + c)])}]
 1/Sqrt[a b c d]  1/4 Sum[(-1)^(Floor[(eta - 1)/2]) (-1)^
   Floor[(xi - 1)/2] FF[0, Sqrt[(b + c)/(a + b + c)]/(
    1 + Sqrt[a/(
     a + b + c)]), (-1)^Floor[(xi - 1)/2] I Sqrt[d/(b + c)] + (-1)^(
      xi - 1) I Sqrt[( b + c + d)/(
      b + c)], -(-1)^(1 + eta + 
        Floor[(eta - 1)/2]) I Exp[(-1)^(Floor[(eta - 1)/2]) I ArcTan[
        Sqrt[c]/Sqrt[b]]]], {xi, 1, 4}, {eta, 1, 4}]

Actualización: como un control de cordura, mire el caso$a=b=c=d=1$. Luego define:

$$\begin{eqnarray} M1&:=&\left( \begin{array}{cccc} -1+\sqrt{3} & \sqrt{2} & \sqrt{2} & -1+\sqrt{3} \\ 1 & \sqrt{2-\sqrt{3}} & \sqrt{2-\sqrt{3}} & 1 \\ \sqrt{2-\sqrt{3}} & 1 & 1 & \sqrt{2-\sqrt{3}} \\ \sqrt{2} & -1+\sqrt{3} & -1+\sqrt{3} & \sqrt{2} \\ \end{array} \right)\\ M2&:=&\left( \begin{array}{cccc} \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) \\ \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) \\ \end{array} \right)\\ A1&:=&\left( \begin{array}{cccc} -\frac{\pi }{6} & \frac{\pi }{12} & -\frac{\pi }{4} & 0 \\ \frac{5 \pi }{6} & \frac{\pi }{12} & \frac{5 \pi }{12} & -\frac{\pi }{3} \\ -\frac{5 \pi }{12} & \frac{\pi }{3} & -\frac{5 \pi }{6} & -\frac{\pi }{12} \\ \frac{\pi }{4} & 0 & \frac{\pi }{6} & -\frac{\pi }{12} \\ \end{array} \right)\\ A2&:=&\left( \begin{array}{cccc} -\frac{\pi }{12} & \frac{\pi }{6} & -\frac{\pi }{6} & \frac{\pi }{12} \\ \frac{7 \pi }{12} & -\frac{\pi }{6} & \frac{\pi }{6} & -\frac{7 \pi }{12} \\ -\frac{\pi }{6} & \frac{7 \pi }{12} & -\frac{7 \pi }{12} & \frac{\pi }{6} \\ \frac{\pi }{6} & -\frac{\pi }{12} & \frac{\pi }{12} & -\frac{\pi }{6} \\ \end{array} \right) \end{eqnarray}$$ y tenemos $$\begin{eqnarray} I(1,1,1,1)=\frac{1}{4} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \left( Li_2(M1_{\xi,\eta}\exp(\imath A1_{\xi,\eta}))- Li_2(M2_{\xi,\eta}\exp(\imath A2_{\xi,\eta})) \right) \end{eqnarray}$$ Hemos comprobado numéricamente que esta cantidad anterior coincide con$\pi^2/96$a cien dígitos. Sería interesante probar esto analíticamente.

Esta integral es hasta la integral constante de normalización de la distribución gaussiana multivariada. Debido a la falta de términos cruzados, obtenemos que hay 4 variables gaussianas independientes de media cero involucradas.

Podemos comenzar reescribiendo esto como una probabilidad de un evento.

Dejar$X,Y,Z,W$independientes distribuidas normalmente con media cero y varianzas posiblemente diferentes.

Entonces la integral se reduce a:

$$\mathbb{P} (0 < W < Z < Y < X)$$

Que a su vez es lo mismo que:

$$\mathbb{P}(W > 0 \land Z-W >0 \land Y-Z >0 \land X-Y >0)$$

y uno puede observar que$W,Z-W,Y-Z,X-Y$son variables conjuntamente normales correlacionadas. La probabilidad de que todos los componentes de un vector normal conjunto sean positivos se denomina probabilidad ortante y, en general, no tiene expresión de forma cerrada. Supongo que este es un caso bastante especial con una matriz de covarianza casi diagonal, por lo que tal vez haya algunos artículos sobre cómo abordar este caso especial.

Para el caso de 3 o menos variables, se conocen fórmulas (ver esta pregunta, por ejemplo) y supongo que coincidirían con lo que has encontrado.

Aquí damos una respuesta usando un método diferente. Asumir que$a\ge0$,$b\ge 0$,$c\ge 0$y$d\ge 0$. Definir:

$$\begin{equation} I(a,b,c,d):=\int _0^{\infty }\int _0^x\int _0^y\int _0^z \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2}-\frac{d w^2}{2} \right) \,\mathrm{d}w\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \end{equation}$$ Entonces nosotros tenemos: $$\begin{eqnarray} &&I(a,b,c,d)=\\ &&\sqrt{\frac{\pi}{2 c d}} \int\limits_0^\infty \int\limits_0^x \int\limits_0^{\sqrt{c} y}\exp(-\frac{a x^2}{2} - \frac{b y^2}{2}-\frac{ z^2}{2})\cdot \text{erf}(\sqrt{\frac{d}{2 c}} z)\; dz \; dy \;dx=\\ &&\frac{2 \pi}{\sqrt{c d}} \int\limits_0^\infty \int\limits_0^x \exp(-\frac{a x^2}{2}-\frac{b y^2}{2}) \left(\frac{1}{2\pi} \arctan[\sqrt{\frac{d}{c}}] - T(\sqrt{c} y,\sqrt{\frac{d}{c}}\right)\; dy \; dx=\\ && \frac{2 \pi^2}{\sqrt{a b c d}}\int\limits_0^\infty \frac{\exp(-\frac{y^2}{2})}{\sqrt{2\pi}} \text{erfc}(\sqrt{\frac{a}{2 b}y})\left(\frac{1}{2\pi} \arctan[\sqrt{\frac{d}{c}}] - T(\sqrt{\frac{c}{b}} y,\sqrt{\frac{d}{c}}\right) \; dy=\\ && \frac{2 \pi^2}{\sqrt{a b c d}} \left( \frac{\left(\pi -2 \tan ^{-1}\left(\sqrt{\frac{a}{b}}\right)\right) \tan ^{-1}\left(\sqrt{\frac{d}{c}}\right)-\pi \sin ^{-1}\left(\frac{\sqrt{b d}}{\sqrt{(b+c) (c+d)}}\right)}{4 \pi ^2}+ \int\limits_0^\infty \frac{\exp(-\frac{y^2}{2})}{\sqrt{2 \pi}} \text{erf}(\sqrt{\frac{a}{2 b}} y) T(\sqrt{\frac{c}{b}} y,\sqrt{\frac{d}{c}} )\; dy \right)=\\ && \frac{2 \pi^2}{\sqrt{a b c d}} \left( \frac{\tan ^{-1}\left(\sqrt{\frac{d}{c}}\right)-\tan ^{-1}\left(\frac{\sqrt{b d}}{\sqrt{c (b+c+d)}}\right)}{4 \pi } +\right.\\ &&\left. \frac{1}{8 \pi^2} \sum\limits_{i=1}^4 \sum\limits_{j=1}^4 (-1)^{\left\lfloor \frac{i-1}{2}\right\rfloor +j} {\mathfrak F}^{(1,\sqrt{\frac{a+b+c}{b+c}}-\sqrt{\frac{a}{b+c}})}_{i \left((-1)^{\left\lfloor \frac{j-1}{2}\right\rfloor } \sqrt{\frac{b+c+d}{b+c}}+(-1)^j \sqrt{\frac{d}{b+c}}\right),(-1)^{i+1} \sqrt{\frac{c}{b+c}}+i \sqrt{\frac{b}{b+c}} (-1)^{\left\lceil \frac{i-1}{2}\right\rceil +1}} \right) \end{eqnarray}$$ En la primera línea integramos sobre$w$utilizando la definición de la función de error. En la segunda línea integramos sobre$z$utilizando la definición de la función T de Owen. En la tercera línea intercambiamos el orden de integración e integramos sobre$x$usando la definición de la función de error complementaria. En la cuarta línea dividimos la integral en integrales factibles y una más complicada y finalmente en la quinta línea evaluamos la integral restante usando una integral que involucra una Gaussiana, funciones de error y la función T de Owen. .

Clear[F]; Clear[FF];
F[z_, a_, b_] := 
  Log[a + z] Log[(b + z)/(-a + b)] + PolyLog[2, (a + z)/(a - b)];
FF[A_, B_, a_, b_] := 
  Module[{result, ts, zs, zsp, zsm, eps = 10^(-50)},
   (*This is Integrate[Log[z+a]/(z+b),{z,A,B}] where all a,b,A, 
   and B are complex. *)
   result = F[B, a, b] - F[A, a, b];


   ts = - (Im[(A + b) (Conjugate[b] - Conjugate[a])]/
     Im[(B - A) (Conjugate[b] - Conjugate[a])]);
   If[0 <= ts <= 1,
    zsp = A + (ts + eps) (B - A);
    zsm = A + (ts - eps) (B - A);
    result += -F[zsp, a, b] + F[zsm, a, b];
    ];

   result
   ];
J[a_, b_, c_] := 
  1/ Pi^2 (ArcTan[Sqrt[2] a]/2 ArcTan[ c] + 
     1/8  Sum[
       FF[1, ( Sqrt[1 + 2 a^2 + b^2] - Sqrt[2] a)/Sqrt[
         1 + b^2], ((-1)^j I b c + (-1)^Floor[(j - 1)/2] I Sqrt[
           1 + b^2 + b^2 c^2])/Sqrt[
         1 + b^2], -(((-1)^Ceiling[(i - 1)/2] I + (-1)^i b)/Sqrt[
          1 + b^2])] (-1)^(j + Floor[(i - 1)/2]), {i, 1, 4}, {j, 1, 
        4}] );

{a, b, c, d} = RandomReal[{0, 10}, 4, WorkingPrecision -> 50];
NIntegrate[
 Exp[-a/2 x^2 - b/2 y^2 - c/2 z^2 - d/2 w^2], {x, 0, Infinity}, {y, 0,
   x}, {z, 0, y}, {w, 0, z}]
Sqrt[\[Pi]/2]/(Sqrt[c] Sqrt[d])
  NIntegrate[
  Exp[-a/2 x^2 - b/2 y^2 - 1/2 z^2] Erf[Sqrt[d/(2 c)] z], {x, 0, 
   Infinity}, {y, 0, x}, {z, 0, Sqrt[c] y}]
 (2 \[Pi])/Sqrt[c d]
  NIntegrate[
  Exp[-a/2 x^2 - b/2 y^2] (ArcTan[Sqrt[d/c]]/(2 \[Pi]) - 
     OwenT[Sqrt[c] y, Sqrt[d/c]]), {x, 0, Infinity}, {y, 0, x}]
(2  \[Pi]^2)/Sqrt[a b c d]
  NIntegrate[
  Exp[ -1/2 y^2]/Sqrt[2 Pi]
    Erfc[Sqrt[a/(2 b)] y] (ArcTan[Sqrt[d/c]]/(2 \[Pi]) - 
     OwenT[Sqrt[c/b] y, Sqrt[d/c]]), {y, 0, Infinity}]
(2  \[Pi]^2)/Sqrt[
 a b c d] ((-\[Pi] ArcSin[Sqrt[b d]/
      Sqrt[(b + c) (c + d)]] + (\[Pi] - 2 ArcTan[Sqrt[a/b]]) ArcTan[
      Sqrt[d/c]])/(4 \[Pi]^2) + 
   NIntegrate[
    Exp[ -1/2 y^2]/Sqrt[2 Pi]
      Erf[Sqrt[a/(2 b)] y] OwenT[Sqrt[c/b] y, Sqrt[d/c]], {y, 0, 
     Infinity}])
(2  \[Pi]^2)/Sqrt[
 a b c d] ((-ArcTan[Sqrt[b d]/ Sqrt[c (b + c + d)]] + 
    ArcTan[Sqrt[d/c]])/(4 \[Pi]) +  
   1/( 8 Pi^2)
     Sum[FF[1, Sqrt[(a + b + c)/(b + c)] - Sqrt[a/(b + c)], 
       I  ((-1)^j Sqrt[d/(b + c)] + (-1)^Floor[1/2 (-1 + j)] Sqrt[(
           b + c + d)/(b + c)]), (-1)^(i + 1) Sqrt[c/(b + c)] + 
        Sqrt[b/(b + c)] I (-1)^(1 + Ceiling[1/2 (-1 + i)])] (-1)^(
      j + Floor[(i - 1)/2]), {i, 1, 4}, {j, 1, 4}] )

---

Definir la función$\mathcal{I}:\mathbb{R}_{>0}^{4}\rightarrow\mathbb{R}$a través de la integral cuádruple

$$\mathcal{I}{\left(a,b,c,d\right)}:=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{ax^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}.\tag{1}$$

Si observamos cómo esta integral se transforma bajo la sustitución$\left(x,y,z,w\right)\mapsto\left(px,py,pz,pw\right)$para algunos reales positivos fijos pero arbitrarios$p$, obtenemos la siguiente relación de escala:

$$\forall\left(a,b,c,d,p\right)\in\mathbb{R}_{>0}^{5}:\mathcal{I}{\left(a,b,c,d\right)}=p^{4}\mathcal{I}{\left(ap^{2},bp^{2},cp^{2},dp^{2}\right)}.\tag{2}$$

Como tal, en nuestra evaluación general de$\mathcal{I}{\left(a,b,c,d\right)}$será suficiente considerar la$a=1$caso.

---

Para empezar, derivamos algunas fórmulas de integración rápida que serán útiles a continuación.

Para cualquier$p\in\mathbb{R}_{>0}$,

$$\begin{align} \int_{0}^{\infty}\mathrm{d}x\,x^{3}\exp{\left(-\frac{px^{2}}{2}\right)} &=\int_{0}^{\infty}\mathrm{d}y\,\frac{y}{2}\exp{\left(-\frac{py}{2}\right)};~~~\small{\left[x=\sqrt{y}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}z\,\frac{2}{p}\cdot\frac{z}{p}\exp{\left(-z\right)};~~~\small{\left[y=\frac{2z}{p}\right]}\\ &=\frac{2}{p^{2}}\int_{0}^{\infty}\mathrm{d}z\,z\exp{\left(-z\right)}\\ &=\frac{2}{p^{2}}.\tag{3a}\\ \end{align}$$

A continuación, dado$p\in\mathbb{R}_{>0}$y configuración$q:=\sqrt{p}$,

$$\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+pt^{2}\right)^{2}} &=\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+q^{2}t^{2}\right)^{2}}\\ &=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{2x^{2}}{\left(1+x^{2}\right)^{2}};~~~\small{\left[qt=x\right]}\\ &=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan{\left(x\right)}-\frac{x}{1+x^{2}}\right]\\ &=\frac{1}{q^{3}}\left[\arctan{\left(q\right)}-\frac{q}{1+q^{2}}\right]\\ &=\left[\frac{\arctan{\left(q\right)}}{q^{3}}-\frac{1}{q^{2}\left(1+q^{2}\right)}\right]\\ &=\left[\frac{\arctan{\left(\sqrt{p}\right)}}{p\sqrt{p}}-\frac{1}{p\left(1+p\right)}\right].\tag{3b}\\ \end{align}$$

Definición de la función$\mathcal{J}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$por

$$\mathcal{J}{\left(p,q,x\right)}:=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right],\tag{3c}$$

entonces encontramos que para cualquier$\left(p,q,x\right)\in\mathbb{R}_{>0}^{3}$,

$$\begin{align} \mathcal{J}{\left(p,q,x\right)} &=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right]\\ &=\int_{0}^{x}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{qy^{2}}{p\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)};~~~\small{I.B.P.}\\ &~~~~~-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{p\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{1}{p\sqrt{q\left(1+p\right)}}\int_{0}^{x\sqrt{\frac{q}{1+p}}}\mathrm{d}t\,\frac{1}{\left(1+t^{2}\right)};~~~\small{\left[y=t\sqrt{\frac{1+p}{q}}\right]}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{q}}{\sqrt{1+p}}\right)}}{p\sqrt{q\left(1+p\right)}}.\tag{3d}\\ \end{align}$$

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Suponer$\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$.

Empezamos por reducir$\mathcal{I}$a una integral de una sola variable de la siguiente manera:

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{x^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}\\ &~~~~~\times\exp{\left(-\frac{x^{2}}{2}-\frac{bx^{2}t^{2}}{2}-\frac{cx^{2}u^{2}}{2}-\frac{dx^{2}v^{2}}{2}\right)};~~~\small{\left[\left(y,z,w\right)=\left(xt,xu,xv\right)\right]}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\int_{0}^{\infty}\mathrm{d}x\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}},\\ \end{align}$$

y luego,

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}};~~~\small{\left[\left(u,v\right)=\left(tx,ty\right)\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left[1+\left(b+cx^{2}+dy^{2}\right)t^{2}\right]^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\bigg{[}\frac{\arctan{\left(\sqrt{b+cx^{2}+dy^{2}}\right)}}{\left(b+cx^{2}+dy^{2}\right)^{3/2}}\\ &~~~~~-\frac{1}{\left(b+cx^{2}+dy^{2}\right)\left(1+b+cx^{2}+dy^{2}\right)}\bigg{]}\\ &=\int_{0}^{1}\mathrm{d}x\,\mathcal{J}{\left(b+cx^{2},d,x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right].\tag{4}\\ \end{align}$$

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Definir las funciones auxiliares$\mathcal{F}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$y$\mathcal{G}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$por las respectivas integrales

$$\begin{align} \mathcal{F}{\left(b,c,d\right)} &:=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\tag{5a}\\ \end{align}$$

y

$$\begin{align} \mathcal{G}{\left(b,c,d\right)} &:=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}.\tag{5b}\\ \end{align}$$

Suponer$\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$, y establecer$p:=\sqrt{\frac{b}{c}}\land q:=\sqrt{\frac{1+b}{c}}\land r:=\sqrt{\frac{d}{c}}$. A continuación, notando que$0<p<q$, colocar$s:=\frac{p}{q}\land z:=q^{-1}$. Entonces obtenemos las siguientes expresiones para$\mathcal{G}$y$\mathcal{F}$:

$$\begin{align} \mathcal{G}{\left(b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{rx}{\sqrt{q^{2}+x^{2}}}\right)}}{\left(p^{2}+x^{2}\right)\sqrt{q^{2}+x^{2}}}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{q\arctan{\left(\frac{rqy}{\sqrt{q^{2}+q^{2}y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{q^{2}+q^{2}y^{2}}};~~~\small{\left[x=qy\right]}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{1+y^{2}}}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{z}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(s^{2}+y^{2}\right)\sqrt{1+y^{2}}}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{z}{\sqrt{1+z^{2}}}}\mathrm{d}t\,\frac{\arctan{\left(rt\right)}}{s^{2}+\left(1-s^{2}\right)t^{2}};~~~\small{\left[\frac{y}{\sqrt{1+y^{2}}}=t\right]}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{rz}{\sqrt{1+z^{2}}}}\mathrm{d}u\,\frac{r\arctan{\left(u\right)}}{r^{2}s^{2}+\left(1-s^{2}\right)u^{2}};~~~\small{\left[rt=u\right]}\\ &=\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}},\\ \end{align}$$

y

$$\begin{align} \mathcal{F}{\left(b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x\arctan{\left(\sqrt{b+\left(c+d\right)x^{2}}\right)}}{2\left(b+cx^{2}\right)\sqrt{b+\left(c+d\right)x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{b+\left(c+d\right)y}\right)}}{2\left(b+cy\right)\sqrt{b+\left(c+d\right)y}};~~~\small{\left[x^{2}=y\right]}\\ &=\int_{b}^{b+c+d}\mathrm{d}t\,\frac{\arctan{\left(\sqrt{t}\right)}}{2\left(bd+ct\right)\sqrt{t}};~~~\small{\left[y=\frac{t-b}{c+d}\right]}\\ &=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}};~~~\small{\left[\sqrt{t}=u\right]}.\\ \end{align}$$

Por lo tanto, podemos expresar$\mathcal{I}$como

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right]\\ &=\mathcal{F}{\left(b,c,d\right)}-\mathcal{G}{\left(b,c,d\right)}\\ &=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}-\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}\\ &=\frac{1}{\sqrt{bcd}}\int_{\sqrt{\frac{c}{d}}}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\ &~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}};~~~\small{\left[u=x\sqrt{\frac{bd}{c}}\right]}\\ &=\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{d}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\ &~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}.\tag{6}\\ \end{align}$$

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Finalmente, se puede demostrar (ver Apéndice) que la siguiente fórmula de integración es válida para todos$\left(p,z\right)\in\mathbb{R}_{>0}^{2}$:

$$\int_{0}^{z}\mathrm{d}x\,\frac{2\arctan{\left(px\right)}}{1+x^{2}}=\arctan^{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(-\frac{1-p}{1+p}\right)}-\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},\pi-2\arctan{\left(z\right)}\right)},\tag{7}$$

donde la variante de dos variables del dilogaritmo está definida por la representación integral

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$

Dado que cada una de las tres integrales restantes en la última línea de$(6)$anterior se puede evaluar con la fórmula$(7)$, esto en principio completa la derivación. Sin embargo, no veo mucho sentido en pasar por el tedio de escribir la expresión explícita.

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Apéndice.

Definir la función$\mathcal{K}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$a través de la integral definida

$$\mathcal{K}{\left(a,z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}.$$

En el caso especial$a=1$la integral es elemental y tenemos

$$\mathcal{K}{\left(1,z\right)}=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{1+x^{2}}=\frac12\arctan^{2}{\left(z\right)};~~~\small{z\in\mathbb{R}_{>0}}.$$

Suponer$\left(a,z\right)\in\mathbb{R}_{>0}^{2}$, y asumir$a\neq1$. Entonces,$-1<\frac{1-a}{1+a}<1\land\frac{1-a}{1+a}\neq0$.

Colocar$\omega:=2\arctan{\left(z\right)}$, y tenga en cuenta que$0<\omega<\pi\land z=\tan{\left(\frac{\omega}{2}\right)}$.

$$\begin{align} \mathcal{K}{\left(a,z\right)} &=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}\\ &=\int_{0}^{z}\mathrm{d}x\,\frac{1}{1+x^{2}}\int_{0}^{a}\mathrm{d}t\,\frac{x}{1+x^{2}t^{2}}\\ &=\int_{0}^{z}\mathrm{d}x\int_{0}^{a}\mathrm{d}t\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}t\int_{0}^{z}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\ &=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\left(1+t^{2}y\right)};~~~\small{\left[x^{2}=y\right]}\\ &=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{d}{dy}\left[\frac{\ln{\left(1+y\right)}-\ln{\left(1+t^{2}y\right)}}{1-t^{2}}\right]\\ &=\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(1+z^{2}\right)}-\ln{\left(1+z^{2}t^{2}\right)}}{1-t^{2}}\\ &=-\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(\frac{1+z^{2}t^{2}}{1+z^{2}}\right)}}{1-t^{2}}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+z^{2}\left(\frac{1-x}{1+x}\right)^{2}}{1+z^{2}}\right)};~~~\small{\left[t=\frac{1-x}{1+x}\right]}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{\left(1+x\right)^{2}+z^{2}\left(1-x\right)^{2}}{\left(1+z^{2}\right)\left(1+x\right)^{2}}\right)}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+2\left(\frac{1-z^{2}}{1+z^{2}}\right)x+x^{2}}{\left(1+x\right)^{2}}\right)}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\left[\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}-2\ln{\left(1+x\right)}\right]\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\ &~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &=+\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac18\omega^{2}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\ &~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{(-1)\ln{\left(1-2x\cos{\left(\pi-\omega\right)}+x^{2}\right)}}{2x}\\ &=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-a}{1+a},\pi-\omega\right)}.\blacksquare\\ \end{align}$$

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