The following is my current solution $$ \displaystyle \begin{aligned}\int_{0}^{\infty}{\dfrac{\ln\left(1+{\large\frac{1}{x}}\right)}{π^2+\ln^2x}\ \mathrm{d}x}&=\dfrac{1}{π}\int_{-\infty}^{\infty}{\dfrac{e^{πx}\ln(1+e^{-πx})}{1+x^2}} \ \mathrm{d}x\\\\&=\dfrac{1}{π}\left(\int_{0}^{\infty}{\dfrac{e^{πx}\ln(1+e^{-πx})}{1+x^2}\ \mathrm{d}x}-\int_{0}^{\infty}{\dfrac{e^{-πx}\ln(1+e^{πx})}{1+x^2}\ \mathrm{d}x}\right)\\\\&=\dfrac{1}{π}\int_{0}^{\infty}{\dfrac{\ln(1+e^{-πx})(e^{πx}-e^{-πx})}{1+x^2}\ \mathrm{d}x}-\int_{0}^{\infty}{\dfrac{xe^{-πx}}{1+x^2}\ \mathrm{d}x}\\\\&=\dfrac{1}{π}\sum_{k=1}^{\infty}{\dfrac{(-1)^{k+1}}{k}\int_{0}^{\infty}{\dfrac{e^{-(k-1)πx}-e^{-(k+1)πx}}{1+x^2}}\ \mathrm{d}x}-\mathrm{Ci}(π)\\\\&=\dfrac{1}{π}\sum_{k=1}^{\infty}{\dfrac{\mathrm{Si}((k+1)π)-\mathrm{Si}((k-1)π)}{k}}-\mathrm{Ci}(π)\\\\&=-\dfrac{1}{π}\int_{0}^{π}{\dfrac{2\sin x\ln\left(2\sin\left({\large\frac{x}{2}}\right)\right)}{x}\ \mathrm{d}x}-\mathrm{Ci}(π)\\\\&=-\dfrac{2}{π}\int_{0}^{1}{\dfrac{2x\ln\left(2x\right)}{\arcsin x}\ \mathrm{d}x}-\mathrm{Ci}(π)\\\\&=-\dfrac{2}{π}\int_{0}^{1}{\dfrac{2x\ln x}{\arcsin x}\ \mathrm{d}x}-\dfrac{2\ \mathrm{Si}(π)\ln2}{π}-\mathrm{Ci}(π)\end{aligned} $$ Neither the Feynman integral nor the series nor the residue, I can't simplify the integral in the last step. Hasty in waiting for tycoons to provide ideas. With many thanks!
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Nice work for sure and (+1) – Claude Leibovici Jul 08 '23 at 05:00
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By writing the original integral as a double integral $$I = \int_0^\infty \int_0^1 \frac{dy:dx}{(\pi^2+\log^2x)(x+y)}$$ Fubini's theorem tells us $$I = \frac{1}{2} + \int_0^1 f(\log y):dy$$ where $f(\cdot)$ is some odd function. I haven't made much progress on finding out what $f$ is since its derivatives are divergent integrals. – Ninad Munshi Jul 08 '23 at 05:46
2 Answers
$\require{AMScd}$ I actually have a conjecture regarding this integral, it equals to euler-mascheroni constant $\gamma$. Unfortunately, I currently had no ways to verify it other then the shaky conputation power provided by Desmos. I shall post my argument below. \begin{align*} &I = \int_0^{\infty}{\frac{\ln \left( 1+\frac{1}{x} \right)}{\pi ^2+\ln ^2\left( x \right)}\mathrm{d}x}, &I_t \stackrel{\mathrm{def}}{=} \int_0^{\infty}{\frac{\ln \left( 1+\frac{t}{x} \right)}{\pi ^2+\ln ^2\left( x \right)}\mathrm{d}x}\\ &I_1 =I, I_0 =0 \end{align*} Where $t\in[0,1]$. Then, differentiating with respect to $t$ $$ \Rightarrow I'_t =\int_0^{\infty}{\frac{1}{\left( x+t \right) \left[ \pi ^2+\ln ^2\left( x \right) \right]}\mathrm{d}x} $$ I would use some complex analysis in the next part. Consider $$ f\left( z \right) =\frac{1}{\left( z+t \right) \left( \ln \left( z \right) -i\pi \right)} $$ By setting the direction of branch cut along positive real axis, and integral along the keyhole contour $C$, we have \begin{align*} \lim_{\begin{smallmatrix}R\rightarrow \infty\\ \varepsilon \rightarrow 0\end{smallmatrix}}\oint_C{f\left( z \right)}\mathrm{d}z&=\int_{\rightarrow}{f\left( z \right) \mathrm{d}z}+\int_{\varGamma _R}{f\left( z \right) \mathrm{d}z}+\int_{\gets}{f\left( z \right) \mathrm{d}z}+\int_{\gamma _{\varepsilon}}{f\left( z \right) \mathrm{d}z}\\ &=2\pi i\mathop {\mathrm{Res}} \limits_{z=-1}\left[ f\left( z \right) \right] +2\pi i\mathop {\mathrm{Res}} \limits_{z=-t}\left[ f\left( z \right) \right] =\frac{2\pi i}{1-t}+\frac{2\pi i}{\ln \left( t \right)} \end{align*} Now let's caluculate those four integrals individually. To start with, we have \begin{align*} &\left| \int_{\varGamma _R}{f\left( z \right) \mathrm{d}z} \right|\leqslant \int_{\varGamma _R}{\left| f\left( z \right) \right|\mathrm{d}z}\leqslant \frac{2R}{R-t}\left| \frac{\delta \left( \varepsilon \right) -\pi}{\ln \left( R \right) -\pi} \right| \\ &\left| \int_{\gamma _{\varepsilon}}{f\left( z \right) \mathrm{d}z} \right|\leqslant \int_{\gamma _{\varepsilon}}{\left| f\left( z \right) \right|\mathrm{d}z}\leqslant \frac{4\varepsilon}{t-\varepsilon}\ln \left( \frac{\ln \left( \varepsilon \right)}{\ln \left( \varepsilon \right) -\pi +\eta \left( \varepsilon \right)} \right) \end{align*} Where $\delta \left( \varepsilon \right) $ and $\eta \left( \varepsilon \right) $ goes to zero when $\varepsilon$ approaches zero. Then by squeeze theorem, both of the integrals vanishes in the limit. \begin{align*} &\int_{\rightarrow}{f\left( z \right) \mathrm{d}z}+\int_{\gets}{f\left( z \right) \mathrm{d}z} \\ &=\int_{\varepsilon}^R{\frac{1}{\left( x+t \right) \left( \ln \left( x \right) -i\pi \right)}\mathrm{d}x}+\int_R^{\varepsilon}{\frac{1}{\left( x+t \right) \left( \ln \left( x \right) +i\pi \right)}\mathrm{d}x} \\ &=\int_{\varepsilon}^R{\frac{2i\pi}{\left( x+t \right) \left[ \pi ^2+\ln ^2\left( x \right) \right]}\mathrm{d}x} \end{align*} Therefore, \begin{align*} \lim_{\begin{smallmatrix}R\rightarrow \infty\\ \varepsilon \rightarrow 0\end{smallmatrix}}\oint_C{f\left( z \right)}\mathrm{d}z&=2\pi iI'_t=\frac{2\pi i}{1-t}+\frac{2\pi i}{\ln \left( t \right)}\\ &\Rightarrow I'_t=\frac{1}{1-t}+\frac{1}{\ln \left( t \right)} \end{align*} Hence, we have \begin{align*} I&=\int_0^1{I'_t}\mathrm{d}t=\int_0^1{\left( \frac{1}{1-t}+\frac{1}{\ln \left( t \right)} \right)}\mathrm{d}t \\ &=\underset{\varepsilon \rightarrow 1^-}{\lim}\int_0^{\varepsilon}{\left( \frac{1}{1-t}+\frac{1}{\ln \left( t \right)} \right)}\mathrm{d}t=\underset{\varepsilon \rightarrow 1^-}{\lim}\left( -\ln \left( 1-\varepsilon \right) +\int_0^{\varepsilon}{\frac{1}{\ln \left( t \right)}}\mathrm{d}t \right) \\ &\stackrel{\ln \left( t \right) =-x}{\begin{CD}@=\end{CD}}\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \ln \left( 1-\varepsilon \right) -\int_{-\ln \left( \varepsilon \right)}^{\infty}{\frac{e^{-x}}{x}}\mathrm{d}x \right) \\ &=\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \ln \left( 1-\varepsilon \right) -\varepsilon \ln \left( -\ln \left( \varepsilon \right) \right) -\int_{-\ln \left( \varepsilon \right)}^{\infty}{\ln \left( x \right) e^{-x}}\mathrm{d}x \right) \\ &=\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \ln \left( 1-\varepsilon \right) -\varepsilon \ln \left( -\ln \left( \varepsilon \right) \right) \right) -\int_0^{\infty}{\ln \left( x \right) e^{-x}}\mathrm{d}x \end{align*} The limit here, \begin{align*} &\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \ln \left( 1-\varepsilon \right) -\varepsilon \ln \left( -\ln \left( \varepsilon \right) \right) \right) \\ &=\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \ln \left( 1-\varepsilon \right) -\ln \left( -\ln \left( \varepsilon \right) \right) \right) +\underset{\varepsilon \rightarrow 1^-}{\lim}\left( \left( \varepsilon -1 \right) \ln \left( -\ln \left( \varepsilon \right) \right) \right) \\ &=\ln \left( \underset{\varepsilon \rightarrow 1^-}{\lim}\left( \frac{1-\varepsilon}{-\ln \left( \varepsilon \right)} \right) \right) +0\stackrel{\mathrm{L'H}}{=}\ln \left( \underset{\varepsilon \rightarrow 1^-}{\lim}\left( \varepsilon \right) \right) =0 \end{align*} At last, the integral $$ \int_0^{\infty}{\ln \left( x \right) e^{-x}}\mathrm{d}x=\Gamma '\left( 1 \right) =\psi _0\left( 1 \right) =-\gamma $$ So, I arrived at the result that $$I = \int_0^{\infty}{\frac{\ln \left( 1+\frac{1}{x} \right)}{\pi ^2+\ln ^2\left( x \right)}\mathrm{d}x}=\gamma$$
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It is incorrect. Numerically, the integral is $0.551056$, but $\gamma=0.577216$ – MathFail Jul 15 '23 at 23:44
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That's why I said it was shaky... but I wonder where the argument went wrong. By https://math.stackexchange.com/questions/2943416, the result of contour integration should be correct, and the latter part is verified by wolfram alpha. – oO_ƲRF_Oo Jul 16 '23 at 01:05
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This is pretty much correct. (+1) But how does $\left| \int_{\varGamma R}{f\left( z \right) \mathrm{d}z} \right|\leqslant \int{\varGamma R}{\left| f\left( z \right) \right|\mathrm{d}z}$? I would imagine $\Gamma{R}$ is that huge circular arc that almost makes a full circle, so integrating $|f(z)|$ over that arc would return an imaginary value, I would think. A better way to bound your integral is by either using the ML-inequality or parameterizing it by letting $z = Re^{it}$ where $t$ is in some interval that is approximately $[0, 2\pi]$. Same with $\gamma_{\epsilon}$. Also, $t \in (0,1)$. – Accelerator Jul 16 '23 at 09:55
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And by "Same with $\gamma_{\epsilon}$," I mean that you seem to have the same bounding issue with it. And I am assuming $\arg z \in [0,2\pi)$ since you said the branch cut is on the positive real axis. – Accelerator Jul 16 '23 at 09:56
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I'm pulling this from my analysis book: Take any region $E$ in complex plane $f\in L^1(E,\mu), \exists,, \theta ,,\mathrm{s.t.}$ $$ \int_E{f\left( x \right)}\mathrm{d}\mu=e^{i\theta}\left| \int_E{f\left( x \right)}\mathrm{d}\mu \right| $$ – oO_ƲRF_Oo Jul 16 '23 at 12:28
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Therefore.$$ \left| \int_E{f\left( x \right)}\mathrm{d}\mu \right|=\mathrm{Re}\left{ \left| \int_E{f\left( x \right)}\mathrm{d}\mu \right| \right} =\mathrm{Re}\left{ e^{-i\theta}\int_E{f\left( x \right)}\mathrm{d}\mu \right} \ =\int_E{\mathrm{Re}\left{ e^{-i\theta}f\left( x \right) \right}}\mathrm{d}\mu\leqslant \int_E{\left| e^{-i\theta}f\left( x \right) \right|}\mathrm{d}\mu\leqslant \int_E{\left| f\left( x \right) \right|}\mathrm{d}\mu $$ – oO_ƲRF_Oo Jul 16 '23 at 12:30
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As for the bounding issue, can't we just put a indicator function into the integrand? However, I do think ML lemma is a better way to compute it, I was too tired to conceive that... – oO_ƲRF_Oo Jul 16 '23 at 12:34
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(You have to @ me) What do you mean by indicator function? As for your textbook, I can't really follow it. I've always thought that $\left|\int_c f(z) , dz \right| \leq \int_c |f(z)| \cdot |dz|$ but I don't know if your textbook is basically saying the same thing. – Accelerator Jul 16 '23 at 20:17
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@Accelerator No, it's not. It really just means taking absolute value on the value of the integral. I guess it's mostly a problem on my side cause I'd never taken a complex analysis course, yet I enjoy solving integral using residue theorem. :) – oO_ƲRF_Oo Jul 16 '23 at 22:24
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I didn't interpret your textbook to be saying it just takes the absolute value on the value of the integral. The source you provide seems to be saying that you can just bound the absolute value of a function's integral by the integral of the function's absolute value, regardless of what bounds the integral takes. I find that weird because as far as I know, integrating a real-valued expression like $|f(z)|$ over a complex region $E$ could result in a complex number. This would be an issue because complex numbers don't have a canonical ordering. – Accelerator Jul 16 '23 at 22:33
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@Accelerator Oh! Sorry, yeah the way I typed it was misleading. I meant to put the absolute value in the integral after I parameterize the curve. If that make sense. – oO_ƲRF_Oo Jul 16 '23 at 22:36
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Hm, I think I see what you're saying. And that's fine. I make a lot of mistakes in my answers/comments tol. :) – Accelerator Jul 16 '23 at 22:43
Too long for a comment $$I=\int_{0}^{1}\frac{x \log (x)}{\sin ^{-1}(x)}\,dx$$ it could be interesting to consider the $[2n,2n]$ Padé approximant $P_n$ of $\frac{x }{\sin ^{-1}(x)}$ which write $$P_n=\frac {1+\sum_{k=0}^n a_k\, x^{2k} } {1+\sum_{k=0}^n b_k\, x^{2k} }$$ and using partial fraction decomposition to write it as $$P_n=\frac{a_n}{b_n}\,\sum_{k=1}^n \frac {c_k}{x^2-r_k}$$ where (not proved) all $r_k >1$.
This would lead to $$J_n=\int_0^1 P_n\,\log(x)\,dx=\frac{a_n}{b_n}\,\sum_{k=1}^n \frac {c_k}{4 r_k^2} \left(4 r_k+\Phi \left(\frac{1}{r_k},2,\frac{3}{2}\right)\right)$$ where $\Phi(.)$ is the Lerch transcendent function.
Converted to numerical, some results $$\left( \begin{array}{cc} n & J_n \\ 1 & -0.979262 \\ 2 & -0.978741 \\ 3 & -0.978687 \\ 4 & -0.978677 \\ 5 & -0.978674 \\ 6 & -0.978673 \\ \end{array} \right)$$
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