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Few days ago, I met a complicated integral in a youtube $$ I=\int_0^{\infty} \frac{x^2(x-1)}{\left(x^2+1\right)^3 \ln x} d x, $$ I tackled it by transforming it into a double integral $$ \begin{aligned} I&=\int_0^{\infty} \frac{x^2}{\left(x^2+1\right)^3} \int_0^1 x^a d a d x \\ & =\int_0^1\left( \underbrace{ \int_0^{\infty} \frac{x^{a+2}}{\left(x^2+1\right)^3} d x}_{J} \right) d a \end{aligned} $$ Letting $y=\frac{1}{x^2+1}$ transforms $J$ into a beta function $$ \begin{aligned} J & =\frac{1}{2} \int_0^1 y^{\frac{1}{2}-\frac{a}{2}}(1-y)^{\frac{a}{2}+\frac{1}{2}} d y \\ & =\frac{1}{2} B\left(\frac{3-a}{2}, \frac{3+a}{2}\right)\\&=\frac{1}{2} \frac{\Gamma\left(\frac{3-a}{2}\right) \Gamma\left(\frac{3+a}{2}\right)}{\Gamma(3) } \\ & =\frac{1}{4} \cdot \frac{1-a}{2} \cdot \Gamma\left(\frac{1-a}{2}\right) \cdot \frac{1+a}{2}\cdot \Gamma\left(\frac{1+a}{2}\right)\\&=\frac\pi {16}\left(1-a^2\right)\sec\left( \frac{\pi a}{2} \right) \end{aligned} $$ where the last steps use $\Gamma(1+x)=x\Gamma(x)$ and the reflection property of Gamma function.

Now integrating back, $$ \begin{aligned} I & =I(1)-I(0) \\ & = \frac{\pi}{16}\int_0^1\left(1-a^2\right) \sec \left(\frac{\pi a}{2}\right) d a \end{aligned} $$ Via integration by parts, we have $$ \begin{aligned} \int_0^1\left(1-a^2\right) \sec \left(\frac{\pi a}{2}\right) d a = & \frac{2}{\pi} \int_0^1\left(1-a^2\right) d\left(\ln \left(\sec \frac{\pi a}{2}+\tan \frac{\pi a}{2}\right)\right) \\ = & \frac{4}{\pi} \int_0^1 a \ln \left(\sec \frac{\pi a}{2}+\tan \frac{\pi a}{2}\right) d a \\ = & \frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} x \ln (\sec x+\tan x) d x ,\quad \textrm{ where }x=\frac{\pi a}{2} \\=& \frac{16}{\pi^3} \left( \underbrace{ \int_0^{\frac{\pi}{2}}x \ln (1+\sin x)dx}_{K} - \underbrace{ \int_0^{\frac{\pi}{2}} x \ln (\cos x)d x}_{L} \right) \end{aligned} $$

$$ \begin{aligned} K & =\int_0^{\frac{\pi}{2}} x \ln (1+\sin x) d x \\ & =\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right) \ln (1+\cos x) d x \\ & =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (1+\cos x) d x-\int_0^{\frac{\pi}{2}} x \ln (1+\cos x) d x \end{aligned} $$ $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \ln (1+\cos x) d x=&\int_0^{\frac{\pi}{2}} \ln \left(2 \cos ^2 \frac{x}{2}\right) d x \\ = & \int_0^{\frac{\pi}{2}} \ln 2 d x+2 \int_0^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\ = & \frac{\pi}{2} \ln 2+4 \int_0^\frac \pi4 \ln (\cos x) d x \cdots (*)\\ = & \frac{\pi}{2} \ln 2+2 G-\pi \ln 2 \\ = & 2 G-\frac{\pi}{2} \ln 2 \end{aligned} $$ where $(*)$ use the result obtained in the post.

$$ \begin{aligned}\int_0^{\frac{\pi}{2}} x \ln (1+\cos x) d x = & \int_0^{\frac{\pi}{2}} x \ln \left(2 \cos ^2 \frac{x}{2}\right) d x \\ = & \frac{\pi^2}{8} \ln 2 +2 \int_0^{\frac{\pi}{2}} x \ln \left(\cos \frac{x}{2}\right) d x \\ = & \frac{\pi^2}{8} \ln 2 +8 \int_0^{\frac{\pi}{4}} x \ln (\cos x) d x \\ = & \frac{\pi^2}{8} \ln 2 +8 \cdot \frac{1}{128}\left(16 \pi G-21 \zeta(3)-4 \pi^2 \ln 2\right) \\ = & \frac{\pi^2}{8} \ln 2 +\pi G-\frac{21}{16} \zeta(3)-\frac{\pi^2}{4} \ln 2 \\ = & \pi G-\frac{21}{16} \zeta(3)-\frac{\pi^2}{8} \ln 2 \end{aligned} $$ $$ \boxed{K=\frac{21}{16} \zeta(3) -\frac{\pi^2}{8} \ln 2} $$

For the integral $L$, using the result obtained in the post, we have $$ \boxed{L=-\frac{7\zeta(3)} {16} -\frac{\pi^2}{8} \ln 2} $$

Plugging back yields $$ \boxed{I=\frac{7 \zeta(3)}{4 \pi^2}} $$

My Question:

Do we have any simpler method to deal with the integral $I$?

Your comments and alternative methods are highly appreciated.

Lai
  • 31,615
  • 2
    $$\frac{\pi}{16}\int_{0}^{1}(1-x^2)\sec\left(\frac{\pi x}{2}\right)dx=\frac{7\zeta(3)}{4\pi^2}$$ $$\iff\int_{0}^{1}\frac{\arctan(t)\arctan(1/t)}{t}dt=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^3}$$ I only got this far though. Not sure it is helpful. – Quý Nhân Jan 15 '25 at 12:03
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    $$I=\int_0^{\infty} \frac{x^2(x-1)}{\left(x^2+1\right)^3 \ln x} d x\overset{t=\frac1x}{=}\int_0^\infty\frac{t^3(t-1)}{\left(t^2+1\right)^3 \ln t} d t$$ $$\Rightarrow I=\frac12\int_0^{\infty}\frac{(x+x^2)(x-1)}{\left(x^2+1\right)^3 \ln x} d x\overset{x=e^t}{=}\frac18\int_{-\infty}^\infty\frac{\sinh t}{\cosh^3t}\frac{dt}{t}\overset{IBP}{=}\frac1{16}\int_{-\infty}^\infty\frac{\tanh^2t}{t^2}dt$$ – Svyatoslav Jan 15 '25 at 13:56
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    Closing the contour in the upper half-plane $$=\frac{2\pi i}{16}\sum\underset{z=\frac{\pi i}2+\pi ik}{\operatorname{Res}}\frac{\tanh^2z}{z^2}=\frac{32\pi}{16\pi^3}\sum_{k=0}^\infty\frac1{(1+2k)^3}=\frac2{\pi^2}\zeta(3)\left(1-\frac1{2^3}\right)=\frac{7\zeta(3)}{4\pi^2}$$ – Svyatoslav Jan 15 '25 at 14:09
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    @Svyatoslav. You should post it as an answer. – Abezhiko Jan 15 '25 at 14:30

3 Answers3

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$$ \begin{align*} \mathcal{I} &= \frac{\pi}{16} \int_0^1 (1-a^2) \sec \left(a \frac{\pi}{2} \right) \, \mathrm{d}a \\& \overset{\text{IBP}}{=} \frac{1}{8} \int_0^1 a \ln \left(\frac{1+\sin\left(\frac{a \pi}{2}\right)}{1-\sin\left(\frac{a \pi}{2}\right) } \right) \, \mathrm{d}a \\&\overset{\frac{a \pi}{2}=u}{=} \frac{1}{2\pi^2} \int_0^{\frac{\pi}{2}} u \ln \left(\frac{1+\sin(u)}{1-\sin(u)}\right) \, \mathrm{d}u \\&= \frac{-1}{\pi^2} \int_0^{\frac{\pi}{2}} u \ln \left( \tan\left(\frac{\pi}{4}-\frac{u}{2}\right) \right) \, \mathrm{d}u \\&= \frac{1}{\pi^2} \int_0^{\frac{\pi}{2}} \left(u-\frac{\pi}{2}\right) \ln \left( \tan \left(\frac{u}{2} \right) \right) \, \mathrm{d}u \end{align*} $$

Now we apply the Fourier series of $\ln(\tan(x))$ to obtain

$$ \begin{align*} \mathcal{I} &= \frac{2}{\pi^2} \sum_{n=0}^{\infty} \frac{1}{2n+1} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2}-u\right) \cos((2n+1)u) \, \mathrm{d}u \\&= \frac{2}{\pi^2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^3} \\&= \frac{7 \zeta(3)}{4 \pi^2} \end{align*} $$

We conclude that

$$\int_0^{\infty} \frac{x^2(x-1)}{(x^2+1)^3 \ln(x)} \, \mathrm{d}x= \frac{7 \zeta(3)}{4\pi^2} $$

antas
  • 440
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Differentiate $\int_0^\infty \frac{\sin(t s)}{\sinh\frac{\pi s}2 } ds = \tanh t$ twice to get $$\int_0^\infty \frac{s^2\sin(t s)}{\sinh\frac{\pi s}2 } ds = 2\tanh t \ \text{sech}^2 \ t$$ Then \begin{align} &\int_0^{\infty} \frac{x^2(x-1)}{\left(x^2+1\right)^3 \ln x}\overset{x\to \frac1x}{ d x} = \frac12\int_0^{\infty} \frac{x(x^2-1)}{\left(x^2+1\right)^3 \ln x} \overset{t=\ln x}{ d x} \\ =& \ \frac14 \int_{0}^{\infty}\frac{\tanh{t}\ \text{sech}^2 \ t}{t}dt = \frac18\int_{0}^{\infty}\int_0^\infty \frac{s^2 \sin(t s)}{t\sinh\frac{\pi s}2 } ds\ dt\\ = &\ \frac\pi{16}\int_{0}^{\infty} \frac{s^2}{\sinh\frac{\pi s}2 }{ds} \overset{\ln y={-\frac{\pi s}2}} = \frac1{\pi^2} \int_0^1 \frac{\ln^2y}{1-y^2}dy =\frac7{4\pi^2}\zeta(3) \end{align}

Quanto
  • 120,125
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$$ \begin{aligned} \int_0^1 \frac{1-x^2}{\cos \left(\frac{\pi x}{n}\right)} d x = & 2 \int_0^1 \frac{1-x^2}{e^{\frac{\pi xi}{2}}+e^{-\frac{\pi xi}{2}}} d x \\ = & 2 \int_0^1 \frac{e^{-\frac{\pi}{2} x i}\left(1-x^2\right)}{1+e^{-\pi x i}} d x \\ = & 2 \sum_{n=0}^{\infty}(-1)^n \int_0^1\left(1-x^2\right) e^{-\left(n+\frac{1}{2}\right) \pi x i} d x \\ = &2 \sum_{n=0}^{\infty}(-1)^n \int_0^1\left(1-x^2\right) \cos \left(\left(n+\frac{1}{2}\right) \pi x\right) d x \quad \textrm{(As I is real.) } \\ = & 16 \sum_{n=0}^{\infty}(-1)^n \frac{\pi(2 n+1) \sin (\pi n)+2 \cos (\pi n)}{\pi^3(2 n+1)^3} \quad \textrm{ (By IBP twice)} \\ = & \frac{32}{\pi^3} \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^3} \\ = & \frac{28 \zeta(3)}{\pi^3} \end{aligned} $$

Hence we have $$ \boxed{I =\frac{\pi}{16} \cdot \frac{28 \zeta(3)}{\pi^3} =\frac{7 \zeta(3)}{4\pi^2}} $$

Lai
  • 31,615