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How can I solve this integral? $$I=\int_0^\infty \frac{\ln(2x)^2 \cosh x} {1+\cosh 2x} dx$$ I get this: $I=I_1+I_2$ and $I_1=\int_0^\infty\frac{\ln 2}{e^x+e^{-x}}dx=\pi/2$ and $I_2=\int_0^\infty\frac{\ln x}{e^x+e^{-2x}}dx$ and I can't solve $I_2$. Any help would be appreciated.

Lorago
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Costas
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  • You’re neglecting the square of the logarithm, and your evaluation of $I_1$ is wrong – FShrike Feb 11 '22 at 18:09
  • Did you mean $(\ln 2x)^2$ or $\ln((2x)^2)$? – FShrike Feb 11 '22 at 18:11
  • Is $I_2$'s integrand's denominator definitely meant to be $e^x+e^{-\color{red}{2}x}$? – J.G. Feb 11 '22 at 18:11
  • yeah its:$\int_0^\infty\frac{2ln2xcoshx}{cosh^2x}dx= \int_0^\infty\frac{ln2x}{e^x+e^{-x}}dx= \int_0^\infty\frac {ln2}{e^x+e^{-x}}dx +\frac {lnxe^{-x}}{1+e^{-2x}}dx $ – Costas Feb 12 '22 at 12:59

2 Answers2

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Notice$$\int_0^\infty\frac{x^{s-1}\cosh x}{1+\cosh2x}dx=\int_0^\infty\frac{x^{s-1}e^{-x}dx}{1+e^{-2x}}=\Gamma(s)\beta(s),$$with $\beta$ the Dirichlet beta function, so$$\int_0^\infty\frac{x^{s-1}\ln^nx\cosh x}{1+\cosh2x}dx=\frac{d^n}{ds^n}(\Gamma(s)\beta(s)).$$You need to evaluate the cases $n=0$ and $n=1$ (and also the case $n=2$, if $\ln(2x)^2$ meant $(\ln(2x))^2$) at $s=1$. You'll find $\beta(1),\,\beta^\prime(1)$ here. But if anyone knows $\beta^{\prime\prime}(1)$, they've got Wolfram Alpha beat.

J.G.
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After @J.G.'s answer $$\beta(1)=\frac \pi 4$$ $$\beta'(1)=-\frac{1}{4} \left(\gamma _1\left(\frac{1}{4}\right)-\gamma _1\left(\frac{3}{4}\right)+2 \pi \log (2)\right)$$ $$\beta''(1)=\frac{1}{4} \left(\gamma _2\left(\frac{1}{4}\right)-\gamma _2\left(\frac{3}{4}\right)+4 \left(\gamma _1\left(\frac{1}{4}\right)-\gamma _1\left(\frac{3}{4}\right)\right) \log (2)+4 \pi \log ^2(2)\right)$$

If I am not wrong $$I_2=\frac{1}{18} \left(-3 \log ^2(2)+3 \log (2) \log (3)+\sqrt{3} \pi \left(\log \left(\frac{128}{27}\right)+8 \log (\pi )-12 \log \left(\Gamma \left(\frac{1}{3}\right)\right)\right)\right)$$

Claude Leibovici
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