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This question is from the MIT Integration Bee 2023 Semifinal 2 and is Question 3. The goal is to show $$\int_{0}^{\infty} \frac{\tanh(x)}{x\cosh(2x)}\,\textrm{d}x = \log(2)$$

Ideally, this should be done in three minutes or less. Something I tried doing at first was Feynman's trick, but I ran into a cancellation when trying to evaluate $$I^\prime(t) = \int_{0}^{\infty} \operatorname{sech}^2(tx)\operatorname{sech}(2x)\,\textrm{d}x$$ in that I'd get something along the lines of $I^\prime(t) = -I^\prime(t)$ in the end. Is my approach correct, or is there something that allows me to ignore the hyperbolic functions?

Stamp
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  • Would solving the problem answer your question? – Kamal Saleh Feb 21 '23 at 17:51
  • Solving the equation at the end and integrating gives $I(t) = c$, and I'm pretty sure that isn't the right approach to the question considering the correct answer involves logs. It's either that, or I did my Feynman's trick wrong; I had set $$I(t) = \int_{0}^{\infty} \frac{\tanh(tx)}{x\cosh(2x)},\textrm{d}x$$ – Stamp Feb 21 '23 at 18:06
  • Lol i swear it’s almost always frunalli integral spam – Captain Chicky Feb 21 '23 at 20:18

1 Answers1

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Note $$ \frac{\tanh(x)}{\cosh(2x)}=\frac2{1+e^{2x}}-\frac2{1+e^{4x}}. $$ Let $$ f(x)=\frac2{1+e^{x}} $$ and then $$\int_{0}^{\infty} \frac{\tanh(x)}{x\cosh(2x)}\,\textrm{d}x =\int_0^\infty\frac{f(2x)-f(4x)}x\,\textrm{d}x=(f(\infty)-f(0))\ln(\frac24)=\ln2 $$ by Frullani's Integral

xpaul
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