# Finding $\lim_{a\to \infty} \left(\int\limits_0^\infty\frac {\ln ax}{\cosh x}\, dx-\frac {\pi\ln a}2 \right)$

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...

• @pisco125 It was a mistake on my part. It should be fixed now – Crescendo Jul 25 '17 at 4:03
• 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. – pisco Jul 25 '17 at 4:29
• @pisco125 Okay, is it possible to integrate using Feynman's trick then? – Crescendo Jul 25 '17 at 21:36
• Unlikely this can be done using this trick. – 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 $$I = \frac{\pi}{2}\left[2\ln 2 + 3\ln \pi -4 \ln\Gamma(\frac{1}{4}) \right]$$