Evaluating Series Integral

Question

How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$

I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.

Answer 1

Let $f(x)=\frac{\ln^n(x)}{1+x^2}$

$$I_n=\int_0^\infty f(x)\ dx=\int_0^1f(x)\ dx+\underbrace{\int_1^\infty f(x)\ dx}_{1/x\to x}=\int_0^1f(x)\ dx+\int_0^1(-1)^nf(x)\ dx$$

Clearly, for odd $n$, $I_n=0$, so for even $n$ , we have

$$I_n=2\int_0^1\frac{\ln^n(x)}{1+x^2}\ dx=2\sum_{k=0}^\infty(-1)^k\int_0^1 x^{2k}\ln^n(x)\ dx$$ $$=2(-1)^nn!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{n+1}}=2(-1)^nn!\beta(n+1)$$

and since $n$ is even, we better set $n=2m$

$$\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\ dx=2(2m)!\beta(2m+1),\quad m=0,1,2,...$$

where $\beta(s)$ is The Dirichlet beta function.

Answer 2

Here is a way to evaluate that integral without using the Residue theorem :

We know : $ \left(\forall z\in\mathbb{C}\setminus\mathbb{Z}_{-}\right),\ \Gamma\left(z\right)=\displaystyle\int_{0}^{+\infty}{t^{z}\mathrm{e}^{-t}\,\mathrm{d}t}=2\displaystyle\int_{0}^{+\infty}{u^{2z-1}\mathrm{e}^{-u^{2}}\,\mathrm{d}u} \cdot $

Let $ n\in\mathbb{N}^{*} $ :

\begin{aligned} \left(\forall z\in\mathbb{C}\setminus\mathbb{Z}\right),\ \Gamma\left(z\right)\Gamma\left(1-z\right) &=\left(2\displaystyle\int_{0}^{+\infty}{y^{2z-1}\mathrm{e}^{-y^{2}}\,\mathrm{d}y}\right)\left(2\displaystyle\int_{0}^{+\infty}{x^{1-2z}\mathrm{e}^{-x^{2}}\,\mathrm{d}x}\right) \\ &=4\displaystyle\int_{0}^{+\infty}{\displaystyle\int_{0}^{+\infty}{\left(\displaystyle\frac{y}{x}\right)^{2z-1}\mathrm{e}^{-\left(x^{2}+y^{2}\right)}\,\mathrm{d}x}\,\mathrm{d}y} \\ &=4\displaystyle\int_{0}^{\frac{\pi}{2}}{\displaystyle\int_{0}^{+\infty}{\tan^{2z-1}{\theta}\,\mathrm{e}^{-r^{2}}r\,\mathrm{d}r}\,\mathrm{d}\theta} \\ &=4\left(\displaystyle\int_{0}^{+\infty}{r\mathrm{e}^{-r^{2}}\,\mathrm{d}r}\right)\left(\displaystyle\int_{0}^{\frac{\pi}{2}}{\tan^{2z-1}{\theta}\,\mathrm{d}\theta}\right) \end{aligned} Using the substitution : $ \left\lbrace\begin{aligned}u &=\tan{\theta} \\ \mathrm{d}\theta &=\frac{\mathrm{d}u}{1+u^{2}}\end{aligned}\right. $ we get : \begin{aligned} \left(\forall z\in\mathbb{C}\setminus\mathbb{Z}\right),\ \Gamma\left(z\right)\Gamma\left(1-z\right)&=2\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{u^{2z-1}}{1+u^{2}}\,\mathrm{d}u} \end{aligned} Another substitution : $ \left\lbrace\begin{aligned}u &=t^{\frac{n}{2}} \\ \mathrm{d}u &=\frac{n}{2}t^{\frac{n}{2}-1} \end{aligned}\right. $ gives the following : \begin{aligned} \left(\forall z\in\mathbb{C}\setminus\mathbb{Z}\right),\ \Gamma\left(z\right)\Gamma\left(1-z\right)&=n\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{u^{nz-1}}{1+u^{n}}\,\mathrm{d}u} \end{aligned}

Thus : $$ \fbox{$ \begin{array}{rcl}\left(\forall n\in\mathbb{N}^{*}\right)\left(\forall z\in\mathbb{C}\setminus\mathbb{Z}\right),\ n\displaystyle\int_{0}^{+\infty}{\frac{x^{nz-1}}{1+x^{n}}\,\mathrm{d}x}=\displaystyle\frac{\pi}{\sin{\pi z}}\end{array} $} $$

Let $ n\in\mathbb{N}^{*} : $

Define the function $ f_{n} $, $ \left(\forall t\in\mathbb{R}\setminus\overline{n-1}\right) $ as : $ f_{n}\left(t\right)=\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{x^{t}}{1+x^{n}}\,\mathrm{d}x} \cdot $

Using the previous formula, we get : $ \left(\forall t\in\mathbb{R}\setminus\overline{n-1}\right),\ f_{n}\left(t\right)=\displaystyle\frac{1}{n}\Gamma\left(\displaystyle\frac{t+1}{n}\right)\Gamma\left(1-\displaystyle\frac{t+1}{n}\right)=\displaystyle\frac{\pi}{n\sin{\left(\frac{t+1}{n}\pi\right)}} $

Hence : $$ \fbox{$ \begin{array}{rcl}\left(\forall n\in\mathbb{N}^{*}\right)\left(\forall t\in\mathbb{R}\setminus\overline{n-1}\right),\ \displaystyle\int_{0}^{+\infty}{\displaystyle\frac{x^{t}}{1+x^{n}}\,\mathrm{d}x}=\displaystyle\frac{\pi}{n\sin{\left(\frac{t+1}{n}\pi\right)}}\end{array} $} $$

Observe : $ \left(\forall t\in\mathbb{R}\setminus\overline{n-1}\right),\ f_{n}''\left(t\right)=\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{\partial^{2}}{\partial t^{2}}\left(\displaystyle\frac{x^{t}}{1+x^{n}}\right)\mathrm{d}x}=\displaystyle\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\left(\displaystyle\frac{\pi}{n\sin{\left(\frac{t+1}{n}\pi\right)}}\right) $

Thus : $$ \fbox{$ \begin{array}{rcl}\left(\forall n\in\mathbb{N}^{*}\right)\left(\forall t\in\mathbb{R}\setminus\overline{n-1}\right),\ \displaystyle\int_{0}^{+\infty}{\displaystyle\frac{x^{t}\ln^{2}{x}}{1+x^{n}}\,\mathrm{d}x}=\displaystyle\frac{\pi^{3}\left(1+\cos^{2}{\left(\frac{t+1}{n}\pi\right)}\right)}{n^{3}\sin^{3}{\left(\frac{t+1}{n}\pi\right)}}\end{array} $} $$

Conclusion : Setting $ t=0 $ et $ n=2 $, gives in particular : $$ \fbox{$ \begin{array}{rcl} \displaystyle\int_{0}^{+\infty}{\displaystyle\frac{\ln^{2}{x}}{1+x^{2}}\,\mathrm{d}x}=\displaystyle\frac{\pi^{3}}{8}\end{array} $} $$

Answer 3

Starting with Integrand's observation in the comments, we have $$ \begin{array}{rcl} \displaystyle I = 2\int_{0}^{1}{\displaystyle\frac{\log^{2}{x}}{1+x^{2}}\,\mathrm{d}x}=2 \int_0^1 \log^2{x} \sum_{k \ge 0} (-1)^k x^{2k}\,\mathrm{d}x = 2\sum_{k \ge 0} (-1)^k\int_0^1 x^{2k}\log^2{x}\,\mathrm{d}x\end{array} $$

Now, consider $\displaystyle f(\alpha) = \int_0^1 x^{2\alpha}\,\mathrm{dx} = \frac{1}{1+2\alpha}$. Then finding $f''(\alpha)$ for both sides gives:

$$\displaystyle \int_0^1 x^{2\alpha} \log^2{x}\,\mathrm{dx} = \frac{2}{(1+2\alpha)^3}$$

Therefore:

$$I = 4 \sum_{k \ge 0} \frac{(-1)^k}{(2k+1)^3} = 4 \cdot \frac{\pi^3}{32} = \frac{\pi^3}{8}. $$

Answer 4

\begin{align*} J&=\int_0^\infty \frac{\ln^2 x}{1+y^2}\,dx\\ A&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &=\int_0^\infty \int_0^\infty \frac{\ln^2(x)+\ln^2(y)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &=\int_0^\infty \int_0^\infty \frac{2\ln^2(x)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &=2J\Big[\arctan x\Big]_0^\infty\\ &=\pi J \end{align*} On the other hand, \begin{align*} A&\overset{u(x)=xy}=\int_0^\infty \left(\int_0^\infty \frac{\ln^2 u}{y(1+\left(\frac{u}{y}\right)^2)(1+y^2)}\,du\right)\,dy\\ &=\int_0^\infty \left(\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}\,du\right)\,dy\\ &=\frac{1}{2}\int_0^\infty \left[\frac{\ln\left(\frac{u^2+y^2}{1+y^2}\right)}{1-u^2}\right]_{y=0}^{y=\infty}\ln^2 u\,du\\ &=\int_0^\infty \frac{\ln^3 u}{u^2-1}\,du\\ &=\int_0^1 \frac{\ln^3 u}{u^2-1}\,du+\int_1^\infty \frac{\ln^3 x}{x^2-1}\,dx\\ &\overset{u=\frac{1}{x}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}\,du\\ &=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\int_0^1 \frac{2x\ln^3 x}{x^2-1}\,dx\\ &\overset{u=x^2}=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du\\ &=\frac{15}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du \end{align*} Moreover, \begin{align*} \int_0^1 \frac{\ln^3 u}{1-u}\,du&=\int_0^1 \left(\sum_{n=0}^\infty u^n\right)\ln^3 u\,du\\ &=\sum_{n=0}^\infty\left(\int_0^1 u^n\ln^3 u\,du\right)\\ &=-6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\\ &=-6\zeta(4) \end{align*} Therefore, \begin{align*}\ A&=\frac{15}{8}\times 6\zeta(4)\\ &=\frac{45}{4}\zeta(4)\\ J&=\frac{A}{\pi}\\ &=\frac{45\zeta(4)}{4\pi}\\ &=\frac{45\times\frac{\pi^4 }{90}}{4\pi}\\ &=\boxed{\frac{\pi^3}{8}} \end{align*}

NB: Observe that $\displaystyle \int_0^1 \frac{\ln x}{1+x^2}\,dx\overset{u=\frac{1}{x}}=-\int_0^1 \frac{\ln u}{1+u^2}\,du$

and, i assume $\displaystyle \zeta(4)=\dfrac{\pi^4}{90}$