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Context: I looked up "complex residue" on google images, and saw this integral. I, being unfamiliar with the use of contour integration, decided to try proving the result without complex analysis. Seeing as I was stuck, I decided to ask you for help.

I am attempting to prove that $$J=\int_0^\infty\frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}8$$ With real methods because I do not know complex analysis.

I have started with the substitution $x=\tan u$: $$J=\int_0^{\pi/2}\log^2(\tan x)\mathrm dx$$ $$J=\int_0^{\pi/2}\log^2(\cos x)\mathrm dx-2\int_{0}^{\pi/2}\log(\cos x)\log(\sin x)\mathrm dx+\int_0^{\pi/2}\log^2(\sin x)\mathrm dx$$ But frankly, this is basically worse.

Could I have some help? Thanks.


Update: Wait I think I actually found a viable method

$$F(\alpha)=\int_0^\infty \frac{x^{\alpha}}{x^2+1}\mathrm dx$$ As I have shown in other posts of mine, $$\int_0^\infty\frac{x^{2b-1}}{(1+x^2)^{a+b}}\mathrm dx=\frac12\mathrm{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{2\Gamma(a+b)}$$ so $$F(\alpha)=\frac12\Gamma\left(\frac{1+\alpha}2\right)\Gamma\left(\frac{1-\alpha}2\right)$$ And from $$\Gamma(1-s)\Gamma(s)=\frac\pi{\sin \pi s}$$ we see that $$F(\alpha)=\frac\pi{2\cos\frac{\pi \alpha}{2}}$$ So $$J=F''(0)=\frac{\pi^3}8$$

Okay while I have just found a proof, I would like to see which ways you did it.

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    $\begingroup$ Are you familiar with the Beta function? $\endgroup$ – Zachary Jan 29 at 16:59
  • $\begingroup$ @Zachary very. There's an edit coming up which uses it $\endgroup$ – clathratus Jan 29 at 17:02
  • $\begingroup$ The post has a question, but not much context. Is that integral of particular interest? Why are you interested in evaluating it? Where did you encounter it? We have additional information at "How to ask a good question": math.meta.stackexchange.com/questions/9959/… $\endgroup$ – Carl Mummert Jan 29 at 17:03
  • $\begingroup$ @CarlMummert The upcoming edit will include more context. $\endgroup$ – clathratus Jan 29 at 17:05
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    $\begingroup$ I've favourited and upvoted this question because, if you stitch together multiple methods from its answers, you can instead present it as a computation of $\beta(3)$, and serendipitously we've reduced a question about the Dirichlet beta function to properties of the other "Beta" function. $\endgroup$ – J.G. Jan 29 at 17:56
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Pretty straightforward this integral can be related to the Dirichlet Beta Function $\beta(s)$ and its integral representation which is given by

$$\beta(s)~=~\frac1{\Gamma(s)}\int_0^\infty \frac{t^{s-1}}{e^{t}+e^{-t}}\mathrm dt$$

Therefore, enforce the substitution $x=e^{-t}$ within your integral $J$ to obtain

\begin{align*} J=\int_0^\infty\frac{\log^2(x)}{1+x^2}\mathrm dx &= \int_{-\infty}^\infty\frac{t^2}{1+e^{-2t}}e^{-t}\mathrm dt\\ &=\int_0^\infty\frac{t^2}{1+e^{-2t}}e^{-t}\mathrm dt+\underbrace{\int_{-\infty}^0\frac{t^2}{1+e^{-2t}}e^{-t}\mathrm dt}_{t~\mapsto~ -t}\\ &=\int_0^\infty\frac{t^2}{e^t+e^{-t}}\mathrm dt+\int_0^\infty\frac{t^2}{1+e^{2t}}e^t\mathrm dt\\ &=2\int_0^\infty\frac{t^2}{e^t+e^{-t}}\mathrm dt\\ &=2\Gamma(3)\beta(3) \end{align*}

$$\therefore~J~=~2\int_0^\infty\frac{t^2}{e^t+e^{-t}}\mathrm dt~=~\frac{\pi^3}8$$

The result follows from the known value $\beta(3)=\frac{\pi^3}{32}$ for which a proofcan be found here for instance. The validity of the integral representation can be shown by utilizing the geometric series of the denominator combined with the change of summation and integration, which is allowed in this case, and followed by applying the Defintion of the Dirichlet Beta Function and the Gamma Function.

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    $\begingroup$ I have never seen this approach before! Thanks for the great answer (+1) $\endgroup$ – clathratus Jan 29 at 17:23
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    $\begingroup$ @clathratus Thank you too! Actually the Riemann Zeta Function, and its relatives, all posses similiar but quite useful integral representations which may be used to evaluate integrals of these types in a closed-form. $\endgroup$ – mrtaurho Jan 29 at 17:26
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    $\begingroup$ @Zacky Honetsly speaking it is quite the same procedure. The only thing one have to do beforehand is to rewrite the integral as $$\int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}\mathrm dt=\int_0^\infty t^{s-1}\frac{e^{-t}}{1+e^{-2t}}\mathrm dt=\int_0^\infty t^{s-1}\left[\sum_{n=0}^\infty e^{-(2n+1)t}\right]\mathrm dt$$ From hereon proceed similiar to the proof for the integral representation of the Riemann Zeta Function, i.e. change the order of integration and summation, substitute $(2n+1)t\mapsto t$ and finish by applying the defintions of the Dirichlet Beta Function and the Gamma Function. $\endgroup$ – mrtaurho Jan 29 at 17:30
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    $\begingroup$ I have accepted your answer as it provides some really interesting relationships to the Dirichlet beta function. Really great work. I wish I could upvote your answer more than once. $\endgroup$ – clathratus Jan 29 at 18:29
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    $\begingroup$ @FDP Although your solution was truly magnificent, I selected this answer because it provided not only a solution but a connection to more general methods of closed forms for similar integrals. I still really appreciate your contribution to this post. $\endgroup$ – clathratus Jan 29 at 20:01
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You can also write it as: $$I=\int_0^1\frac{\log^2(x)}{x^2+1}dx+\int_1^\infty \frac{\log^2(x)}{x^2+1}dx$$ In the second one do a $x\rightarrow \frac{1}{x}$ $$\Rightarrow I=2\int_0^1 \frac{\ln^2 x}{1+x^2}dx =2\sum_{n=0}^\infty (-1)^n\int_0^1 x^{2n}\ln^2 xdx$$$$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^k tdt=\frac{d^k}{da^k} \left(\frac{1}{a+1}\right)=\frac{(-1)^k k!}{(a+1)^{k+1}}$$ $$\Rightarrow I=4\sum_{n=0}^\infty \frac{(-1)^{n} }{(2n+1)^3}=4\cdot\frac{\pi^3}{32}=\frac{\pi^3}{8}$$

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    $\begingroup$ I like this a lot! Thanks Zachy $\endgroup$ – clathratus Jan 29 at 17:18
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    $\begingroup$ Well done (+1). I tried to using the geometric series here too but I did not remember to get rid of the infinite border. The substitution $x\mapsto\frac1x$ is nicely used here! $\endgroup$ – mrtaurho Jan 29 at 17:24
  • $\begingroup$ Just curious ... what real analysis method did you use to evaluate the alternating sum of reciprocal odds? $\endgroup$ – Mark Viola Jan 29 at 19:32
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    $\begingroup$ I used: $$(-1)^n\frac{\pi^{2n+1}E_{2n}}{2^{2n+2}(2n)!} = {\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}}$$ Where $E_{n}$ is the $n$-th Euler number. $\endgroup$ – Zacky Jan 29 at 21:01
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\begin{align} J&=\int_0^\infty\frac{\ln^2 x}{x^2+1}\mathrm dx\\ K&=\int_0^\infty\frac{\ln(x)}{x^2+1}\mathrm dx\\ K^2&=\int_0^\infty\frac{\ln x}{x^2+1}\mathrm dx\int_0^\infty\frac{\ln y}{y^2+1}\mathrm dy\\ &=\int_0^\infty \int_0^\infty\frac{\ln x\ln y}{(x^2+1)(y^2+1)}\\ &=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{\ln(xy)^2-\ln^2 x-\ln^2 y}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\ &=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{\ln(xy)^2}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy-\int_0^\infty \int_0^\infty\frac{\ln^2 x}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\ &=\frac{1}{2}\int_0^\infty \frac{1}{y^2+1}\left(\int_0^\infty\frac{\ln(xy)^2}{x^2+1}\mathrm dx\right)\mathrm dy-\frac{\pi}{2}J \end{align}

Perform the change of variable $u=xy$ ($x$ is the variable)

\begin{align}K^2&=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{y\ln^2 u}{(y^2+1)(y^2+u^2)}\mathrm du\mathrm dy-\frac{\pi}{2}J\\ &=\frac{1}{4}\int_0^\infty\frac{\ln^2 u}{1-u^2}\left[\ln\left(\frac{u^2+y^2}{1+y^2}\right)\right]_{y=0}^{y=\infty}\mathrm du-\frac{\pi}{2}J\\ &=-\frac{1}{2}\int_0^\infty\frac{\ln^3 u}{1-u^2}\mathrm du\\ &=-\frac{1}{2}\left(\int_0^1\frac{\ln^3 u}{1-u^2}\mathrm du+\int_1^\infty\frac{\ln^3 u}{1-u^2}\mathrm du\right)-\frac{\pi}{2}J\\ \end{align}

In the second integral perform the change of variable $v=\dfrac{1}{u}$

\begin{align}K^2&=-\int_0^1\frac{\ln^3 u}{1-u^2}\mathrm du-\frac{\pi}{2}J\\ &=\frac{1}{2}\int_0^1\frac{2u\ln^3 u}{1-u^2}\mathrm du-\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J \\\end{align}

In the first integral perform the change of variable $v=u^2$,

\begin{align}K^2&=\frac{1}{16}\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J\\ &=-\frac{15}{16}\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J\\ &=-\frac{15}{16}\times -6\zeta(4)-\frac{\pi}{2}J\\ &=\frac{45}{8}\zeta(4)-\frac{\pi}{2}J\\ \end{align}

Moreover, $K=0$ (perform the change of variable $v=\frac{1}{x}$)

Therefore,

\begin{align}J&=\frac{45}{4\pi}\zeta(4)\\ &=\frac{45}{4\pi}\times \frac{\pi^4}{90}\\ &=\boxed{\frac{\pi^3}{8}} \end{align}

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    $\begingroup$ Woah! That is really cool! Thanks for one of your usual high quality answers. $\endgroup$ – clathratus Jan 29 at 19:55
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Recall the formula $$\int_0^\infty \frac{t^{x}}{1+t^2}dt=\frac{\pi}{2}\sec\frac{\pi x}{2}$$ Differentiating this formula twice with respect to $x$ yields the answer.

To derive this formula, use the following sequence of substitutions, the Beta Function, and the reflection formula of the Gamma Function: $$\begin{align} \int_0^\infty \frac{t^x}{1+t^2}dt &=\frac{1}{2}\int_0^\infty \frac{t^{(x-1)/2}}{1+t}dt \\ &=\frac{1}{2}\int_0^1 t^{(x-1)/2}(1-t)^{(-1-x)/2}dt\tag{1} \\ &=\frac{1}{2}\frac{\Gamma(\frac{1+x}{2})\Gamma(\frac{1-x}{2})}{\Gamma(1)} \\ &=\frac{\pi}{2}\sec\frac{\pi x}{2} \end{align}$$ where $(1)$was obtained by substituting $t\mapsto \frac{t}{1-t}$.

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    $\begingroup$ (+1) Great minds think alike :) $\endgroup$ – clathratus Jan 29 at 17:12

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