Trouble understanding Blagouchine's extensions to the Malmsten integral

Question

$\newcommand{\d}{\,\mathrm{d}}$The great work of Blagouchine massively generalises the Malmsten integral. I am actually familiar with contour integration, but he consistently introduces results from other papers that I am unfamiliar with, so it feels disingenuous to take notes on a proof I can't understand.

I am interested in arriving at the following in a more "simple" manner:

Let $a\ge0$ be real. Then: $$J(a)=\int_0^\infty\frac{\ln(x^2+a^2)}{\cosh(x)}\d x=2\pi\ln\left(\frac{\Gamma\left(\frac{3}{4}+\frac{a}{2\pi}\right)}{\Gamma\left(\frac{1}{4}+\frac{a}{2\pi}\right)}\right)+\pi\ln2\pi$$

As I mentioned, his derivation is not one that I am particularly keen to follow, but the similarity between this integral and Malmsten's integral made me hopeful one could be derived from the other.

Malmsten's Integral: If $-\pi\lt\varphi\lt\pi$, then:$$I(\varphi)=\int_1^\infty\frac{\ln(\ln(x))}{x^2+2\cdot\cos(\varphi)x+1}\d x=\frac{\pi}{2}\csc(\varphi)\ln\left((2\pi)^{\varphi/\pi}\frac{\Gamma\left(\frac{1}{2}+\frac{\varphi}{2\pi}\right)}{\Gamma\left(\frac{1}{2}-\frac{\varphi}{2\pi}\right)}\right)$$

This I have seen a good proof of and am satisfied with.

Sadly I did not get very far with tackling $J$:

$$J(a)=2\int_0^\infty\frac{\ln(x^2+a^2)}{e^x+e^{-x}}\d x$$Which under the transformation $x\mapsto e^x$ becomes: $$J(a)=2\int_1^\infty\frac{\ln(\ln^2(x)+a^2)}{x^2+1}\d x$$Which is almost equal to $2\cdot I\left(\frac{\pi}{2}\right)$. We want the term inside the logarithm to be simply $\ln(x)$. For clarity, introduce $t$ a dummy variable. For $\ln^2(x)+a^2=\ln(t)$, one gets $t=e^{a^2}x^{\ln(x)}$, and$\d t=2e^{a^2}x^{\ln(x)}\frac{\ln(x)}{x}\d x$, and an expression for $x$ in terms of $t$ too ugly to type.

I do not see how to go any further.

I also do not see how to reverse engineer Blagouchine's closed form into an expression involving $I$ and $a$. It is similar to $I(\pi/2 + a)$, but not close enough.

Must I resign myself to researching his contour integration, or can this be done using Malmsten's integral as a first principle? If it can be done, I am not expecting the full proof, just an indication. Of course, if an answerer wants to derive it fully anyway for the joy of it, feel free!

Answer 1

I fully agree with you that the generalising work of Iaroslav V. Blagouchine (here) allows to easily evaluate a very wide class of integral; particularly, integrals of the forms $\int_{-\infty}^\infty R(x)\ln(x^2+a^2) dx$, where $R(x)$ enjoys the symmetry $R(x)=R(x+2\pi i)$.

There are two general ways of evaluation:

• using the series approach (similar to that used by Carl Malmsten) - the detailed example of such evaluation for the similar integral you can find in this post
• the contour integration

In my opinion, the contour integration provides the shortcut to the answer, saves time and allows to use the original symmetry of the problem. It is not complicated in fact and is based on three main points:

• $\Gamma(1+x)=x\Gamma(x)\,\Rightarrow\,\, \ln x=\ln\Gamma(x+1)-\ln\Gamma (x)$
• $\cosh(x+2\pi i)=\cosh(x)$
• $\Gamma(x)$ does not have zeros (and, therefore, $\ln\Gamma(x)$ does not have branch points) inside the chosen contour in the complex plane

Let's consider the integral $$I(a)=\int_0^\infty\frac{\ln(x^2+a^2)}{\cosh x}dx=2\pi \int_0^\infty\frac{\ln((2\pi t)^2+a^2)}{\cosh 2\pi t}dt$$ $$=2\pi\int_0^\infty\frac{2\ln2\pi}{\cosh2\pi t}dt+2\pi\int_0^\infty\frac{\ln\big(t^2+\frac{a^2}{(2\pi)^2}\big)}{\cosh2\pi t}dt$$ $$=2\ln2\pi\int_0^\infty\frac{dx}{\cosh x}+2\pi\Re\int_{-\infty}^\infty\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t}dt=I_1+I_2$$ $$I_1=2\ln2\pi\int_0^\infty\frac{dx}{\cosh x}=2\ln2\pi\int_{-\infty}^\infty\frac{e^x\,dx}{e^{2x}+1}=\pi\ln2\pi$$ To evaluate $I_2$ we can write $$\ln\big(\frac{a}{2\pi}-it\big)=\ln\Gamma\big(\frac{a}{2\pi}-it+1\big)-\ln\Gamma\big(\frac{a}{2\pi}-it\big)=\ln\Gamma\big(\frac{a}{2\pi}-i(t+i)\big)-\ln\Gamma\big(\frac{a}{2\pi}-it\big)$$ and present the second integral $I_2$, using also the property $\cosh(x+2\pi i)=\cosh(x)$, in the form $$I_2=2\pi\Re\int_{-\infty}^\infty\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t}dt=-2\pi\Re\bigg(\int_{-\infty}^\infty\frac{\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t}dt-\int_{-\infty}^\infty\frac{\ln\Gamma\big(\frac{a}{2\pi}-i(t+i)\big)}{\cosh 2\pi (t+i)}dt\bigg)$$ But we can see that the expression in the parentheses is the integral of the function $\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z}$ along the following contour:

To close the contour we have to add the paths $[1]$ and $[2]$ (it can be shown that integrals along these paths $\to0$ as $R\to\infty$). We have two simple poles inside the contour; therefore $$I_2=-2\pi\,\Re\oint\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z}dz=-2\pi\,\Re\Big(2\pi i\operatorname*{Res}_{\binom{z= i/4}{z=3 i/4}}\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z}\Big)$$ To evaluate the residues we notice (for example, for $z=\frac{i}{4}+\epsilon; \,\epsilon\to0$) $$\frac{1}{\cosh2\pi (\frac{i}{4}+\epsilon)}=\frac{2}{i(e^{2\pi\epsilon}-e^{-2\pi\epsilon})}\to\frac{1}{2\pi i\epsilon}$$ $$I_2=-2\pi\,\Re\Big(2\pi i\frac{\ln\Gamma\big(\frac{a}{2\pi}-i\frac{i}{4}\big)}{2\pi i}-2\pi i\frac{\ln\Gamma\big(\frac{a}{2\pi}-i\frac{3i}{4}\big)}{2\pi i}\Big)=2\pi\frac{\ln\Gamma\big(\frac{a}{2\pi}+\frac{3}{4}\big)}{\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{4}\big)}$$ Taking all together, $$I(a)=I_1+I_2=\pi\ln2\pi+2\pi\frac{\ln\Gamma\big(\frac{a}{2\pi}+\frac{3}{4}\big)}{\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{4}\big)}$$