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When I encountered the integral $\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x $, I tried the substitution $x\mapsto \frac{1}{x} $ and found a wonderful result. $$I=\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x \stackrel{x\mapsto\frac{1}{x}}{=} \int_0^{\infty} \frac{1}{x(1+x)\left(\pi+\ln ^{2 } x\right)} d x $$ Averaging them yields

$$ \begin{aligned} I & =\frac{1}{2} \int_0^{\infty}\left[\frac{1}{(1+x)\left(\pi+\ln ^2 x\right)}+\frac{1}{x(1+x)\left(\pi+\ln ^2 x\right)}\right]dx\\ & =\frac{1}{2} \int_0^{\infty} \frac{1}{x\left(\pi+\ln ^2 x\right)} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{1}{\pi+\ln ^2 x} d(\ln x) \\ & =\frac{1}{2 \sqrt{\pi}}\left[\tan ^{-1}\left(\frac{\ln x}{\sqrt{\pi}}\right)\right]_0^{\infty}\\ & =\frac{\sqrt{\pi}}{2} \end{aligned} $$


Then I want to generalize the result to the integral

$$I_n=\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2n } x\right)} d x $$ By replacing the power $2$ by $2n$, we have $$ \begin{aligned} I_n & =\frac{1}{2} \int_0^{\infty} \frac{1}{x\left(\pi+\ln ^{2 n} x\right)} d x\\ &= \frac{\sqrt[2 n]{\pi}}{2\pi} \int_{-\infty}^{\infty} \frac{1}{1+x^{2 n}} d x \quad (\textrm{ via } \ln x \mapsto \sqrt[2 n]{\pi} x) \end{aligned} $$ Using the well-known result $\int_0^{\infty} \frac{d x}{1+x^n}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$, we have $$ I_n=\frac{\sqrt[2 n]{\pi}}{2 n} \csc \left(\frac{\pi}{2 n}\right) $$ For simplicity, given any even integer $m$, we have $$ \boxed{\int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{m } x\right)} d x =\frac{\sqrt[m]{\pi}}{m} \csc \left(\frac{\pi}{m}\right)} $$


For examples, $$ \begin{aligned} & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{2 } x\right)} d x =\frac{\sqrt{\pi}}{2} \\ & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{4 } x\right)} d x =\frac{\sqrt[4]{\pi}}{2 \sqrt{2}} \\ & \int_0^{\infty} \frac{1}{(1+x)\left(\pi+\ln ^{6 } x\right)} d x =\frac{\sqrt[6]{\pi}}{3} \end{aligned} $$


Are there any other methods? Comments and alternative methods are highly appreciated.

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    $\begingroup$ This is a very elegant solution. (+1) $\endgroup$ Commented Mar 31, 2023 at 11:54
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    $\begingroup$ Your appreciation supports me to do more investigations in future. Thank you very much! $\endgroup$
    – Lai
    Commented Mar 31, 2023 at 12:37
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    $\begingroup$ I am not the type to give compliments for nothing. During the last three years, I followed your questions and answers. You have a real talent. Cheers :-) $\endgroup$ Commented Mar 31, 2023 at 12:43
  • $\begingroup$ Hi Lai, could you tell me what you meant by averaging I and what is the motivation behind that. Great solution by the way... $\endgroup$ Commented Mar 31, 2023 at 12:44
  • $\begingroup$ It means to take the average of the two versions of the integral so as to cancel and hence get rid of the term (1+x) in the denominator. Wish you understand! $\endgroup$
    – Lai
    Commented Mar 31, 2023 at 13:33

1 Answer 1

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Here is a more general extension

$$I_n=\int_0^{\infty} \frac{1}{x(1+x^a)\left(\pi+\ln ^{2n } x\right)}\ dx =\frac{\pi^{\frac1{2n}}}{2n\sin\frac\pi{2n}}$$

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  • $\begingroup$ $I=\frac{1}{2} \int_0^{\infty} \frac{d x}{x^{2-a}\left(\pi+\ln ^{2n }x\right)}$ can’t be continued. Am I right? $\endgroup$
    – Lai
    Commented Apr 26, 2023 at 23:35
  • $\begingroup$ @Lai - You would have $\int_0^{\infty} \frac{dx}{x(1+x^a)\left(\pi+\ln ^{2n } x\right)} \overset{x\to \frac1x}= \int_0^{\infty} \frac{dx}{x(1+x^{-a})\left(\pi+\ln ^{2n } x\right)} $ $\endgroup$
    – Quanto
    Commented Apr 27, 2023 at 0:01
  • $\begingroup$ Yes, you are right! I forgot that you added $x$ in the denominator in addition to $x^a$. Thank you very much for your extension, $\endgroup$
    – Lai
    Commented Apr 27, 2023 at 0:12

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