Can one evaluate Serret's integral using contour integration?

Question

$$\int_0^1 \frac{\ln(x+1)}{x^2+1} dx$$ This is the integral, and if possible could someone tell me whether we could solve any such type of problem via contour integration.

Answer 1

Yes, it is possible to evaluate this integral by means of complex integration (the problem was placed the other day on AoPS site; there are also several other interesting approaches). Probably, complex integration is not a shortcut in this case, but it has some beauty, too, disclosing the symmetry of the integral.

Applying basic transformations to the integral,
$$I=\int_0^1 \frac{\ln(1 + x)}{1 + x^2} dx\overset{x=\frac{1}{t}}{=}\int_1^\infty\frac{\ln(1+t)}{1+t^2}dt-\int_1^\infty\frac{\ln t}{1+t^2}dt=\int_1^\infty\frac{\ln(1+x)}{1+x^2}dx\,-\,G$$ ($G$ is Catalan' constant). $$2I+G=\int_0^\infty\frac{\ln(1+x)}{1+x^2}dx\tag{1}$$ Let's consider $\oint_C\frac{(1+z)^s}{1+z^2}dz\,(s>0)\,$ along the keyhole contour with the cut $[-1;\infty)$, counter-clockwise (we added small circle around $z=-1$ and a big circle - to close the contour; these integrals $I_{r, R}$ tend to zero at $r\to0; R\to\infty$).

On the one hand we have $$\oint_C\frac{(1+z)^s}{1+z^2}dz=I_{r,R}+\, (1-e^{2\pi is})\int_0^\infty\frac{(1+x)^s}{1+x^2}dx+(1-e^{2\pi is})\int_{-1}^0\frac{(1+x)^s}{1+x^2}dx$$ $$=I_{r,R}+\,(1-e^{2\pi is})\int_0^\infty\frac{(1+x)^s}{1+x^2}dx+(1-e^{2\pi is})\int_0^1\frac{(1-x)^s}{1+x^2}dx\tag{2}$$ On the other hand, we have two simple poles inside our closed contour at $z_1=e^{\frac{\pi i}{2}}, z_2=e^{\frac{3\pi i}{2}}$ $$\oint_C\frac{(1+z)^s}{1+z^2}dz=2\pi i\underset{z=z_{1,2}}{\operatorname{Res}}\frac{(1+z)^s}{1+z^2}\tag{3}$$ From (2) and (3), leading $r\to0; R\to\infty$, we get $$(1-e^{2\pi is})\int_0^\infty\frac{(1+x)^s}{1+x^2}dx+(1-e^{2\pi is})\int_0^1\frac{(1-x)^s}{1+x^2}dx=2\pi i\underset{z=z_{1,2}}{\operatorname{Res}}\frac{(1+z)^s}{1+z^2}$$ $$\int_0^\infty\frac{(1+x)^s}{1+x^2}dx+\int_0^1\frac{(1-x)^s}{1+x^2}dx=\frac{2\pi i}{1-e^{2\pi is}}\underset{z=z_{1,2}}{\operatorname{Res}}\frac{(1+z)^s}{1+z^2}\tag{4}$$ Taking the derivative with respect to s, leading $s\to 0$ and using (1), $$\int_0^\infty\frac{\ln(1+x)}{1+x^2}dx+\int_0^1\frac{\ln(1-x)}{1+x^2}dx$$ $$=2I+G+\int_0^1\frac{\ln(1-x)}{1+x^2}dx=\frac{\partial}{\partial s}\bigg|_{s=0}\frac{2\pi i}{1-e^{2\pi is}}\underset{z=z_{1,2}}{\operatorname{Res}}\frac{(1+z)^s}{1+z^2}\tag{5}$$ Now, making the substitution in the second integral $x=\frac{1}{t}$, $$\int_0^1\frac{\ln(1-x)}{1+x^2}dx=\int_1^\infty\frac{\ln(x-1)}{1+x^2}dx-\int_1^\infty\frac{\ln x}{1+x^2}dx=\int_0^\infty\frac{\ln x}{x^2+2x+2}dx-G\tag{6}$$ Using the same approach as above, and a keyhole contour with the cut $[0;\infty)$, it is straightforward to show that $$\int_0^\infty\frac{\ln x}{x^2+2x+2}dx=\frac{\partial}{\partial s}\bigg|_{s=0}\int_0^\infty\frac{x^s}{x^2+2x+2}dx=\frac{\partial}{\partial s}\bigg|_{s=0}\frac{2\pi i}{1-e^{2\pi is}}\underset{z=-1+z_1; -1+z_2}{\operatorname{Res}}\frac{z^s}{z^2+2z+2}$$ $$=\frac{\partial}{\partial s}\bigg|_{s=0}\frac{2\pi i}{1-e^{2\pi is}}\underset{z=z_1; z_2}{\operatorname{Res}}\frac{(-1+z)^s}{z^2+1}\tag{7}$$ Putting (7) and (6) into (5), we see that G cancel, and we finally get: $$\boxed{\,\,2I=\frac{\partial}{\partial s}\bigg|_{s=0}\left(\frac{2\pi i}{1-e^{2\pi is}}\underset{z=e^{\frac{\pi i}{2}}; e^{\frac{3\pi i}{2}} }{\operatorname{Res}}\frac{(1+z)^s-(-1+z)^s}{z^2+1}\right) \,\,}$$ Decomposing RHS into the series (at $s\to 0$) , $$2I\to\,\frac{\partial}{\partial s}\bigg|_{s=0}-\,\frac{1}{2i}\left(1+\pi is\right)\bigg(\ln(1+e^{\frac{\pi i}{2}})-\ln(-1+e^{\frac{\pi i}{2}})-\ln(1+e^{\frac{3\pi i}{2}})+\ln(-1+e^{\frac{3\pi i}{2}})+\frac{s}{2}\Big(\ln^2(1+e^{\frac{\pi i}{2}})-\ln^2(-1+e^{\frac{\pi i}{2}})-\ln^2(1+e^{\frac{3\pi i}{2}})+\ln^2(-1+e^{\frac{3\pi i}{2}})\Big)\bigg)$$ To simplify the evaluation, we can consider only the real part of RHS (our integral is real), thought it is not difficult to check directly that the imaginary terms cancel.

A straightforward evaluation gives $$2I=-\frac{1}{4i}2\ln(\sqrt2)\,\frac{\pi i}{4}\left(1-3-7+5\right)\quad\Rightarrow\quad I=\frac{\pi \ln 2}{8}$$