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how to integrate$$\int_{0}^{1} \int_{0}^{1}\int_0^1 \frac{x^{4a - 1} \ln(x)}{\sqrt{yz} \cdot (1 + x^{2a}z + yzx^{4a} + yx^{2a})} \,dx \,dy\,dz$$

My attempt

$x^{2a} \rightarrow x$

$$=\frac{1}{4a^2} \int_0^1 \int_0^1 \int_0^1 \frac{x \ln(x)}{\sqrt{yz} \cdot (1 + yx)(1 + zx)} \,dx \,dy \,dz$$

$$=\frac{1}{4a^2} \int_0^1 \int_0^1 \int_0^1 \sum_{n, m \geq 0} (-1)^{n+m} y^{n-\frac{1}{2}} z^{m-\frac{1}{2}} x^{n+m+1} \ln(x) \,dx$$

$$= \int_0^1 \sum_{n \geq 0} \frac{(-1)^n x^{n + \frac{1}{2}}}{2n + 1} \cdot \sum_{m \geq 0} \frac{(-1)^m x^{m + \frac{1}{2}} \cdot \ln(x)}{2m + 1} \, dx$$

$$= \frac{1}{a^2} \int_{0}^{1} [\arctan(\sqrt{x})]^2 \ln(x) \,dx$$

$\sqrt{x} = x$

$$=\frac{4}{a^2}\int_{0}^{1} x \ln(x) \arctan^2(x) \,dx$$

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    $\begingroup$ How do you know this integral can be calculated with closed form expression? $\endgroup$
    – NN2
    Feb 12 at 13:02
  • $\begingroup$ @NN2 I don't know the closed form, and I don't think there's an easy proof. $\endgroup$
    – user1285841
    Feb 13 at 6:12

2 Answers 2

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Starting from where OP left: $$\begin{align} I=&\frac{4}{a^2}\int_{0}^{1}x\log{(x)}\arctan^2{(x)}dx\overbrace=^{IBP}\frac{2}{a^2}\underbrace{\left[\left(x^2+1\right)\log{(x)}\arctan^2{(x)}\right]_0^1}_{0}\\ &-\frac{2}{a^2}\int_{0}^{1}\left(\underbrace{x\arctan^{2}( x)}_{I_{1}}+\underbrace{\frac{\arctan^{2}( x)}{x}}_{I_{2}}+2\underbrace{\log( x)\arctan( x)}_{I_{3}}\right)dx \end{align}$$

Integrals $I_1$ and $I_3$ are straight foward, thus I'll not elaborate much about them.Nonetheless, here's a sketch of how to solve them.

$I_1$ is just a matter of integration by parts;adding $1$ as a costant after the first integration (simirlarly as performed above) makes the next computations easier. $$I_1=-\frac{2}{a^{2}}\left(\frac{\pi ^{2}}{16} -\frac{\pi }{4} +\frac{\log( 2)}{2}\right)$$ $I_3$ can be evaluated integrating by parts, but it will require the function $\text{Li}_2(z)$[1]. Alternatively, you may expand $\arctan(x)$ as a series and integrate termwise. $$I_3=-\frac{4}{a^2}\left( -\frac{\pi }{4} +\frac{\log( 2)}{2} +\frac{\pi ^{2}}{48}\right)$$ $I_2$ is a little bit more tricky and will require some algebric manipulation:

$$\begin{align} I_{2} &=-\frac{2}{a^{2}}\int _{0}^{1}\frac{\arctan^{2}( x)}{x} dx=-\frac{2}{a^{2}}\left[\underbrace{\log( x)\arctan^{2}( x)}_{0}\right]_{0}^{1} +\frac{4}{a^{2}}\int _{0}^{1}\frac{\arctan( x)}{1+x^{2}}\log( x) dx\\ =&\frac{2}{a^{2}}\int _{0}^{1}\frac{\arctan( x)}{1+x^{2}}\log( x) dx+\frac{2}{a^{2}}\int _{0}^{1}\underbrace{\frac{\arctan( x)}{1+x^{2}}\log( x)}_{x\rightarrow \frac{1}{x}} dx\\ =&\frac{2}{a^{2}}\left(\int _{0}^{1}\frac{\arctan( x)}{1+x^{2}}\log( x) dx-\int _{1}^{\infty }\frac{\left(\frac{\pi }{2} -\arctan( x)\right)}{1+x^{2}}\log( x) dx\right)\\ =&\frac{2}{a^{2}}\left(\int _{0}^{\infty }\frac{\arctan( x)}{1+x^{2}}\log( x) dx-\frac{\pi }{2}\int _{1}^{\infty }\underbrace{\frac{\log( x)}{1+x^{2}} dx}_{G=\text{Catalan's Constant}}\right)\\ =&\frac{2}{a^{2}}\left( -\frac{\pi G}{2} +\int _{0}^{1} dy\int _{0}^{\infty }\frac{x\log( x)}{\left( 1+x^{2}\right)\left( 1+y^{2} x^{2}\right)} dx\right)\\ =&\frac{2}{a^{2}}\left( -\frac{\pi G}{2} +\frac{1}{2}\int _{0}^{1}\frac{\log^{2}( y)}{1-y^{2}} dy\right)=\frac{2}{a^{2}}\left( -\frac{\pi G}{2} +\sum _{n=0}^{\infty }\frac{1}{( 2n+1)^{3}}\right) =\frac{2}{a^{2}}\left( -\frac{\pi G}{2} +\frac{7\zeta ( 3)}{8}\right) \end{align}$$ Hence: $$I=\frac{4}{a^2}\int_{0}^{1}x\log{(x)}\arctan^2{(x)}dx=\frac{1}{a^{2}}\left[ -\frac{5\pi ^{2}}{24} +\frac{3\pi }{2} -3\log( 2) -\pi G+\frac{7\zeta ( 3)}{4}\right]\\=\boxed{\frac{7\zeta ( 3) -5\zeta ( 2) +2\pi ( 3-2G) -12\log( 2)}{4a^{2}}}$$ [1]: https://mathworld.wolfram.com/Dilogarithm.html

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    $\begingroup$ Thanks a lot for the answer(+1) $\endgroup$
    – user1285841
    Feb 22 at 11:13
  • $\begingroup$ You're welcome! Thanks for the bounty $\endgroup$
    – Teruo
    Feb 22 at 16:29
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$$I=\dfrac1{4a^2}\int\limits_0^1\int\limits_0^1\int\limits_0^1\dfrac{x\ln(x)}{\sqrt{yz}(1+yx)(1+zx)}\text dx\text dy\text dz$$ $$=\dfrac1{4a^2}\int\limits_0^1\int\limits_0^1\int\limits_0^1\dfrac{x\ln(x)}{\sqrt{yz}(1+yx)(1+zx)}\text dy\text dz\text dx=\bigg|y=v^2, z=w^2\bigg|$$ $$=\dfrac1{2a^2}\int\limits_0^1\int\limits_0^1\int\limits_0^1\dfrac{x\ln(x)}{(1+xv^2)(1+xw^2)}\text dv\text dw\text dx$$ $$=\dfrac1{2a^2}\int\limits_0^1\left(\dfrac{\arctan \sqrt x}{\sqrt x}\right)^2 \,x\ln(x)\text dx =\dfrac1{2a^2}\int\limits_0^1 \arctan^2 \left(\sqrt x\right)\ln(x)\text dx$$ $$=\dfrac2{a^2}\int\limits_0^1 \arctan^2 \left(y\right)\ln(y)\, y\text dy$$ $$=\dfrac1{48a^2}((36 - 24\text C - 5\pi)\pi -72\ln2 + 42 ζ(3))\approx -0.098$605673\,\dfrac1{a^2}$$ (see also WA calculations),

where $\text C \approx0.915965594\dots $ is the Catalan's constant.

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    $\begingroup$ Thanks a lot for the answer(+1) $\endgroup$
    – user1285841
    Feb 22 at 11:13

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