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I have created a challenging integral problem, and I dedicate this challenge to fellow integral enthusiasts, inviting them to solve it

$$\small{\int_{0}^{\infty }\frac{\ln\left[\left ( 1+e^{-\sqrt{x}} \right )\left( 1+e^{-3\sqrt{x}}\right)\left( 1+e^{-5\sqrt{x}}\right) \cdots\left ( 1+e^{-\pi ^{2}\sqrt{x}} \right )\left( 1+e^{-3\pi ^{2}\sqrt{x}}\right)\left( 1+e^{-5\pi ^{2}\sqrt{x}}\right)\cdots\right]}{1+x}\mathrm{d}x}$$ $$= \frac{\pi^{3}}{24}+\frac{\pi}{24}-\ln^{2}(2)$$

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    $\begingroup$ Is your integrand $$ \frac{1}{{1 + x}}\ln\left( \prod\limits_{k = 0}^\infty {(1 + e^{ - (2k + 1)\sqrt x } )(1 + e^{ - (2k + 1)\pi ^2 \sqrt x } )}\right) \; ? $$ $\endgroup$
    – Gary
    Commented Sep 22, 2023 at 0:16
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    $\begingroup$ Yes, that is my integrand $\endgroup$
    – SSS
    Commented Sep 22, 2023 at 1:03
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    $\begingroup$ Splitting the logarithm of the product into a sum of logarithms gives: $$I=\int_0^\infty\frac1{1+x}\sum_{k=0}^\infty\left(\ln\left[1+\exp(-(2k+1)\sqrt{x})\right]+\ln\left[1+\exp(-(2k+1)\pi^2\sqrt{x})\right]\right)$$ As the series and integral commute we can swap the order of operations (Fubini's theorem) giving: $$I=\sum_{k=0}^\infty\int_0^\infty\frac{\left(\ln\left[1+\exp(-(2k+1)\sqrt{x})\right]+\ln\left[1+\exp(-(2k+1)\pi^2\sqrt{x})\right]\right)}{1+x}dx$$ Am I on the right track? $\endgroup$
    – Henry Lee
    Commented Sep 22, 2023 at 20:37
  • $\begingroup$ can you post your solution? I did it a very different way. $\endgroup$
    – SSS
    Commented Sep 23, 2023 at 3:25
  • $\begingroup$ @SSS will you post your solution? $\endgroup$ Commented Sep 24, 2023 at 6:11

1 Answer 1

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Let's denote the integral in the OP by $I$, $$I=\int_0^\infty \frac{1}{1+x} \ln\left(\prod_{k=0}^\infty\left(1+e^{-(2k+1)\sqrt{x}}\right)\left(1+e^{-(2k+1)\pi^2\sqrt{x}}\right) \right)dx.$$ Forcing a substitution of $(2k+1)x\to x$ gives \begin{align*} &I=\sum_{k=0}^{\infty} \int_0^{\infty} \frac{2x}{1+x^2} \ln\left((1+e^{-(2k+1)x})(1+e^{-(2k+1) \pi^2 x}\right) dx\\ &=\int_0^{\infty} \left(\ln(1+e^{-x})+\ln(1+e^{-\pi^2 x})\right) \sum_{k=0}^{\infty} \frac{2x}{x^2+(2k+1)^2} dx\\ &=\int_0^{\infty} \left(\ln(1+e^{-x})+\ln(1+e^{-\pi^2 x})\right)\frac{\pi}{2}\text{tanh}(\frac{\pi x}{2}) dx\\ &=\int_0^{\infty} \left(\frac{\pi}{2}\ln(1+e^{-x}) \text{tanh}(\frac{\pi x}{2})+\frac{\pi}{2} \ln(1+e^{- x}) \text{tanh}\big(\frac{\pi \frac{x}{\pi^2}}{2}\big) \frac{1}{\pi^2} \right) dx\\ & = \int_0^\infty \ln(1+e^{-x}) \frac{\pi}{2} \left(\text{tanh}(\frac{\pi x}{2})+\text{tanh}\big(\frac{x}{2\pi}\big) \frac{1}{\pi^2} \right) dx\\ &= \int_0^\infty \frac{1}{e^x+1} \ln\left(\cosh(\frac{\pi x}{2}) \cosh(\frac{x}{2\pi}) \right) dx ,\quad \text{integrate by parts} \\ &= \int_0^\infty \frac{1}{e^x+1} \ln\left(\frac{e^{\frac{x}{2}} (\pi+\frac{1}{\pi})}{4} (e^{-\pi x}+1)(e^{-\frac{x}{\pi}}+1) \right) dx\\ &=\int_0^1 \frac{\ln(x^\pi+1)}{x+1} dx + \int_0^1 \frac{\ln(x^{\frac{1}{\pi}}+1)}{x+1} dx + \int_0^\infty \frac{1}{e^x+1} \left(\frac{x}{2} (\pi+\frac{1}{\pi}) -2 \ln2\right) dx\\ &= \int_0^1 \frac{\ln(x^\pi+1)}{x+1} dx +(\ln2)^2 -\int_0^1 \frac{1}{\pi} x^{\frac{1}{\pi}-1}\frac{\ln(x^{\frac{1}{\pi}}+1)}{x+1} dx +\frac{\Gamma(2) \eta(2)}{2} \left(\pi+\frac{1}{\pi}\right)\\ &\quad -\Gamma(1) \eta(1) 2 \ln2\\ &= \int_0^1 \frac{\ln(x^\pi+1)}{x+1} dx +(\ln2)^2 -\int_0^1 \frac{\ln(x^\pi+1)}{x+1} dx+\frac{\pi}{24} \left(\pi^2+1 \right)-2 (\ln2)^2\\ &= \frac{\pi^3}{24} + \frac{\pi}{24} - (\ln2)^2. \end{align*}

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    $\begingroup$ very nice solution (+1) and welcome to MathSE. That's a good start :) $\endgroup$ Commented Sep 24, 2023 at 5:17
  • $\begingroup$ very good solution $\endgroup$
    – SSS
    Commented Sep 24, 2023 at 6:09

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