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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 ?

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    $\begingroup$ Try doing $u=2\pi t$ to get $(e^{2 \pi t}+1)$ in your denominator and use the Abel -Plana formula to instantly get a series representation for your bottom $\frac12-\int_0^\infty … du$ integral? You need to have $(e^{2 \pi t}-1)$ actually in the denominator to use it. $\endgroup$ Commented Sep 5, 2021 at 1:53
  • $\begingroup$ Related integral:$$\int_{0}^{\pi/2} \frac{x\cot x\ln(\cos x)}{x^2+\ln^2(2\cos x)} \text{d}x -\pi\int_{0}^{\infty} \frac{\frac{1}{x}-\frac{1}{1-e^{-x}}+2-\frac{1}{\ln2} } {e^x-2} \text{d}x =\frac{\pi}{8} \left ( 3+\gamma-7\ln2+3\ln\pi +4\text{Re}\left [ \psi^{(0)}\left ( \frac{\ln2}{2\pi i} \right ) \right ] \right )$$ $\endgroup$ Commented Sep 6, 2021 at 11:59

3 Answers 3

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

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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

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

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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, $$I=\operatorname*{Res}_{z=1}\frac1{(z-1)\log z}=\lim_{z\to 1}\frac{d}{dz}\frac{z-1}{\log z}=\ldots=\frac12.$$

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  • $\begingroup$ I assume the standard CCW orientation of $C_{r,R}$ (unlike the picture on Wikipedia). $\endgroup$
    – metamorphy
    Commented Sep 5, 2021 at 2:47
  • $\begingroup$ Closely related: (1), (2). $\endgroup$
    – metamorphy
    Commented Sep 5, 2021 at 2:58

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