Improper integral $\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$

Question

I am having trouble solving

$$\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$$

let $\ln(x)=u$ then $x du= dx $

$$\Rightarrow \int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx=\int_0^{\infty} \frac{x+1-1}{(x+1)(\pi^2+u^2)}du=\int_0^{\infty} \frac{1}{\pi^2+u^2}du-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$

$$=\frac{1}{2}-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$

How do you proceed from here? I put it into an integral calculator and it stated "Antiderivative or integral could not be found." How do you solve this ?

Answer 1

Let $$I=\int_0^{\infty} \frac{dx}{(x+1)(π^2+(\log x)^2)}.$$

Putting $t=1/x$, we get upon simplification

$$I= \int_0^{\infty} \frac{dt}{t(t+1)(π^2+(\log t)^2)}.$$

Adding the two integrals, we get

$$2I= \int_0^{\infty} \frac{dt}{t(π^2+(\log t)^2}.$$

Substituting $u=\log t$,

$$2I= \int_{-\infty}^{\infty} \frac{du}{π^2+u^2}.$$

Hence, $$I=\frac{1}{2}.$$

Answer 2

Enforcing the substitution $x\mapsto e^{-x}$ reveals

$$\int_0^\infty \frac1{(x+1)(\pi^2+\log^2(x))}\,dx=\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx$$

Then, note that

$$\begin{align} \int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx&=\int_{-\infty}^0 \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\ &=\int_0^\infty \frac{1}{(1+e^{x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\ &=\int_0^\infty \frac{1}{\pi^2+x^2}\,dx\\\\ &=\frac12 \end{align}$$

Answer 3

A complex-analytic approach. Consider $$I=\frac1{2\pi i}\int_{C_{r,R}}\frac{dz}{(z-1)\log z}$$ where $0<r<1<R$, and $C_{r,R}$ is a keyhole contour, say the boundary of $$\big\{z\in\mathbb{C} : r<|z|<R\land(\Re z>0\lor|\Im z|>r)\big\}.$$ As $r\to 0$ and $R\to\infty$, the integrals along $|z|=r$ and $|z|=R$ vanish, hence $$I=\frac1{2\pi i}\left(\int_\infty^0\frac{d(-x)}{(-x-1)(\log x+\pi i)}+\int_0^\infty\frac{d(-x)}{(-x-1)(\log x-\pi i)}\right)$$ is exactly the given integral. On the other hand, by the residue theorem, $$I=\operatorname*{Res}_{z=1}\frac1{(z-1)\log z}=\lim_{z\to 1}\frac{d}{dz}\frac{z-1}{\log z}=\ldots=\frac12.$$