Define the function$$F(a)=\int\limits_0^\infty\frac {\ln ax}{\cosh x}\, dx$$

Question: How do you calculate the limit$$\lim\limits_{a\to\infty}\left(F(a)-\frac {\pi\ln a}2\right)\tag1$$

I was trying to integrate $F(a)$ using Feynman's trick. Differentiating with respect to $a$, we get$$\begin{align*}F'(a) & =\int\limits_{0}^\infty\frac {\text{sech } x}a\\ & =\frac {\pi}{2a}\end{align*}$$Integrating that again, we see that$$F(a)=\frac {\pi\ln a}2+C$$However, in order to find the constant $C$, I need to solve $(1)$. I tried plugging it into Mathematica, and it started running for a long time before I gave up.

I'm wondering if you have any ideas...

  • $\begingroup$ @pisco125 It was a mistake on my part. It should be fixed now $\endgroup$ – Crescendo Jul 25 '17 at 4:03
  • $\begingroup$ Differentiating under integral sign will not work here. Because $\ln ax = \ln a + \ln x$, so the integral can be written as $$(\ln a)*\text{constant} + \text{a function in }x$$ differentiation with respect to $a$ yields nothing about the original integral. $\endgroup$ – pisco Jul 25 '17 at 4:29
  • $\begingroup$ @pisco125 Okay, is it possible to integrate using Feynman's trick then? $\endgroup$ – Crescendo Jul 25 '17 at 21:36
  • $\begingroup$ Unlikely this can be done using this trick. $\endgroup$ – pisco Jul 26 '17 at 5:14

Note that $\ln ax = \ln a + \ln x$, and the following integral (it has elementary antiderivative): $$\int_0^\infty \frac{1}{\cosh x} dx = \frac{\pi}{2}$$

Therefore the original limit equals to: $$I = \lim\limits_{a\to\infty}\left(F(a)-\frac {\pi\ln a}2\right) =\int_0^\infty \frac{\ln x}{\cosh x} dx$$

Perhaps this will suffice as an answer, but this integral can be evaluated using some common functions.

To evaluate it, note that we have the formula $$\int_0^\infty \frac{x^{a-1}}{\cosh x} dx = 2\Gamma(a)\beta(a)$$ where $\beta(a)$ is the Dirichlet beta function. This can be proved by noting $\frac{1}{\cosh x} = \frac{2e^{-x}}{1+e^{-2x}} \quad $and then expand the denominator as a geometric series and integrate termwise.

Hence $$I = 2[\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)]$$ Note that $\Gamma'(1)=-\gamma, \beta(1) = \frac{\pi}{4}$, the only challenge is to calculate $\beta'(1)$. From the definition of $\beta'(1)$, we have $$\beta'(1) = \sum_{n=0}^{\infty} (-1)^n \frac{\ln(2n+1)}{2n+1}$$ This can be evaluate by using Fourier series of $\ln \Gamma(x)$ by plugging in $x=\frac{1}{4}$, I can add more details on this if needed. The result is $$\beta'(1) = \frac{\pi}{4}\left[\gamma + 2\ln 2 + 3\ln \pi -4 \ln\Gamma(\frac{1}{4}) \right] $$

Hence we finally obtain $$I = \frac{\pi}{2}\left[2\ln 2 + 3\ln \pi -4 \ln\Gamma(\frac{1}{4}) \right]$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.