Trouble understanding Blagouchine's extensions to the Malmsten integral

Solution 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, $$I(a)=I_1+I_2=\pi\ln2\pi+2\pi\frac{\ln\Gamma\big(\frac{a}{2\pi}+\frac{3}{4}\big)}{\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{4}\big)}$$