# Table of integrals with Euler's constant gamma.

I'm working on a large table of integrals with Euler's constant $$\gamma=0.577...$$ It's a very interesting point, that a lot of integrals are found in similar tables like Gradshteyn-Ryzhik, but are not able to calculate with maple or mathematica. For example $$\int\limits_0^\infty\frac{\log x}{\cosh^2 x}dx = \log\pi - 2\log2 - \gamma$$ and the similar integral $$\int\limits_0^\infty\log x\left(\frac1{\sinh^2} - \frac1{x^2}\right)dx = \gamma-\log\pi$$ Question 1. It is possible to get the first integral from $$\int\limits_0^\infty\left(\frac1{e^x-1} - \frac1{xe^x}\right)dx = \gamma$$ At StackExchange I found this proof, which could be simplify. Integral $\int_0^{\infty} \frac{\log x}{\cosh^2x} \ \mathrm{d}x = \log\frac {\pi}4- \gamma$ Question 2. Know somebody a proof? It's is easy to get second integral by subtracting both.

Question 3. Are the newest Versions of maple or Mathematica able to do this?

• What's the question? – Jair Taylor Mar 11 at 17:06
• It's not clear what you're asking, and also, aren't those two integrals exactly the same? – Robert Howard Mar 11 at 17:14
• I'm sorry. The entry wasn't finish. – skraemer Mar 11 at 17:35
• So put it in short you want to prove that $$\int\limits_0^\infty\frac{\log x}{\cosh^2 x}dx = \log\pi - 2\log2 - \gamma$$ Using the known result of: $$\int\limits_0^\infty\left(\frac1{e^x-1} - \frac1{xe^x}\right)dx = \gamma$$ Am I right? – Zacky Mar 11 at 17:45
• Yes. But I have found a proof on Stackexange with I have add above. But this proof maybe improved. – skraemer Mar 11 at 17:50