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I have tried many different methods on this particular integral, none of them yielding any fruitful results. Here was an attempt

I(t) = $\int_{0}^{\infty} \frac{ln(tx^{2}+1)} {(x(x^{2}+1))}dx$, I(1) = $\int_{0}^{\infty} \frac{ln(x^{2}+1)}{(x(x^{2}+1))}dx$

I took a partial derivative with respect to $t$, and then integrated with respect to $x$

I'(t) = $\int_{0}^{\infty} \frac{x} {((x^{2}+1)(tx^{2}+1))}dx$ = $\frac{ln(t)} {(2(t-1))}$

This is where I get stuck, because you need to integrate both sides, and then solve an Initial Value Problem to find the missing constant, that's usually how these parametric integrals work out, but I don't believe that integrates exactly. Does anyone have any thoughts on where to proceed?

I also thought of contour integration as way to solve this problem, since this function has decay behavior for large $x$, but I kept getting a cancellation for the integral I was looking for when I was integrating along the real line. The function is odd, so the full real line integral, will just be zero. I'm also not sure how to deal with the branch points for a problem like this. If anyone has any thoughts on a contour integral in the complex plane that could work for this problem, that would really help.

The answer we are looking for is $ \pi^{2}/12$, based on what the answer key says. Any assistance would be greatly appreciated.

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Starting off where you left off: $$I(t)=\int -\frac{1}{2} \cdot \frac{\ln{t}}{1-t} \; dt$$ $$I(t)=-\frac{1}{2} \int \sum_{n=0}^{\infty} t^n \ln{t} \; dt$$ We can interchange the summation and integral sign here because $f(t) \geq 0$ for $0 < t \leq 1$, which is where we are interested in: $$ I(t)=-\frac{1}{2}\sum_{n=0}^{\infty} \int t^n \ln{t} \; dt$$ Using integration by parts with $dv=t^n$ and $u=\ln{t}$: $$ I(t)=-\frac{1}{2}\sum_{n=0}^{\infty} \left( \frac{t^{n+1} \ln{t}}{n+1}-\frac{t^{n+1}}{{(n+1)}^2}+C\right) $$ Note, $I(0)$ is $0$. $$ I(1)=-\frac{1}{2}\sum_{n=0}^{\infty} -\frac{1}{{(n+1)}^2}$$ $$I(1)=\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}$$ Where this is the famous Basel problem or the Riemann zeta function at $s=2$: $$\boxed{I(1)=\frac{\pi^2}{12}}$$

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  • $\begingroup$ Is $I(0)$ well defined? $\endgroup$ – user376343 Jun 5 at 19:56
  • $\begingroup$ Look at $t=0$ in the original integral. Its $tx^2+1$, not $t(x^2+1)$. $I(0)=\int_0^{\infty} \frac{\ln{1}}{x(x^2+1)}=0$ $\endgroup$ – Ty. Jun 5 at 19:57
  • $\begingroup$ thank you guys for you help, and yeah $I(0)=0$ $\endgroup$ – Jessie Christian Jun 5 at 23:54
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One can also find a more generalized version: $$\int _0^{\infty }\frac{\ln \left(x^a+1\right)}{x\left(x^b+1\right)}\:dx$$ $$=\int _0^1\frac{\ln \left(x^a+1\right)}{x\left(x^b+1\right)}\:dx+\underbrace{\int _1^{\infty }\frac{\ln \left(x^a+1\right)}{x\left(x^b+1\right)}\:dx}_{x=\frac{1}{x}}$$ $$=\int _0^1\frac{\ln \left(x^a+1\right)}{x\left(x^b+1\right)}\:dx+\int _0^1\frac{x^{b-1}\ln \left(x^a+1\right)}{x^b+1}\:dx-a\underbrace{\int _0^1\frac{x^{b-1}\ln \left(x\right)}{x^b+1}\:dx}_{x=x^b}$$ $$=\int _0^1\frac{\ln \left(x^a+1\right)}{x}\:dx-\frac{a}{b^2}\int _0^1\frac{\ln \left(x\right)}{x+1}\:dx$$ $$=\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}\int _0^1x^{ak-1}\:dx-\frac{a}{b^2}\sum _{k=0}^{\infty }\left(-1\right)^k\:\int _0^1x^k\:\ln \left(x\right)\:dx$$ $$=\frac{1}{a}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}+\frac{a}{b^2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}$$ $$\boxed{=\eta \left(2\right)\left(\frac{1}{a}+\frac{a}{b^2}\right)}$$ Where $\eta \left(2\right)$ is the dirichlet eta function of $2$.

So, $$\boxed{\int _0^{\infty }\frac{\ln \left(x^a+1\right)}{x\left(x^b+1\right)}\:dx=\eta \left(2\right)\left(\frac{1}{a}+\frac{a}{b^2}\right)}$$ Now for your integral $a=2$,$b=2$. $$\eta \left(2\right)\left(\frac{1}{2}+\frac{2}{4}\right)=\frac{\pi ^2}{12}\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{\pi ^2}{12}$$

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  • $\begingroup$ oh wow that is a wonderful result, thanks so much! $\endgroup$ – Jessie Christian Jun 5 at 23:55

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