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Could this result be proven by methods of differentiation under the integral sign? I only take interest in differentiation way. $$\int_0^\infty \frac{\ln(\tan^2 (ax))}{1+x^2}dx = \pi\ln(\tanh(a))$$

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  • $\begingroup$ I tried to differentiate with respect to $a$ and I got an integral that is impossible to evaluate. $\endgroup$ – Kilgore Trout Apr 28 '13 at 8:35
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    $\begingroup$ Though this is not what you asked for, but my be hlepful. See here and here $\endgroup$ – Norbert Apr 28 '13 at 8:54
  • $\begingroup$ $\ln y^2=2\ln y$ $\endgroup$ – Lucian Nov 17 '13 at 23:22
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Actually using the derivative really simplifies the problem. After that and a simple substitution one obtain: $$ J'(a) = 4 \int_0^\infty \frac{t}{\sin t} \frac{dt}{t^2+4a^2} $$ The last integral can be evaluated by means of complex analysis: $$ J'(a) = \frac{2\pi}{\sinh 2a} $$

I guess this is what they really wanted in this problem.

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