Fourier series for the natural logarithms of trigonometric functions can make quick work of some tricky integrals. In this post we cover two such examples.
The Series
The needed Fourier series are
\[\ln(\sin(x))=-\ln(2)-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\]and
\[\ln(\cos(x))=-\ln(2)-\sum_{n=1}^\infty(-1)^n\frac{\cos(2nx)}{n}.\]The derivation of these is rather short and be found on math.se.
Problem 1
Evaluate
\[I=\int_0^{\frac{\pi}{2}}\frac{x\cdot\sin(x)}{\cos(x)+\sin(x)}\,\mathrm dx.\]We can begin our attack by making the substitution $x\mapsto\frac{\pi}{2}-x$ so that
\[I=\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\cos(x)+\sin(x)}\,\mathrm dx-\int_0^{\frac{\pi}{2}}\frac{x\cdot\cos(x)}{\cos(x)+\sin(x)}\,\mathrm dx.\]Averaging this new representation of $I$ with the original one, we find that
\[I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\cos(x)+\sin(x)}\,\mathrm dx+\frac{1}{2}\int_0^{\frac{\pi}{2}}x\cdot\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)}\,\mathrm dx.\]The $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\cos(x)+\sin(x)}\,\mathrm dx$ integral may be easily evaluated to $\frac{\pi}{4}$ by making the substitution $x\mapsto\frac{\pi}{2}-x$ and doing a similar averaging trick. So
\[I=\frac{\pi^2}{16}+\frac{1}{2}\int_0^{\frac{\pi}{2}}x\cdot\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)}\,\mathrm dx.\]The remaining integral may be dealt with by integrating by parts. Let $u=x$ and $\mathrm dv=\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)}\,\mathrm dx$ so
\[\int_0^{\frac{\pi}{2}}x\cdot\frac{\sin(x)-\cos(x)}{\sin(x)+\cos(x)}\,\mathrm dx=\underset{J}{\underbrace{\int_0^{\frac{\pi}{2}}\ln\left(\cos(x)+\sin(x)\right)\,\mathrm dx}}.\]Then
\[\begin{align*} J &= \int_0^{\frac{\pi}{2}}\ln\left(\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\right)\,\mathrm dx\\ &= \frac{\pi\ln(2)}{4}+\int_0^{\frac{\pi}{2}}\ln\left(\sin\left(x+\frac{\pi}{4}\right)\right)\,\mathrm dx\\ &= \frac{\pi\ln(2)}{4}+\int_0^{\frac{\pi}{2}}\left(-\ln(2)-\sum_{n=1}^\infty\frac{\cos\left(2n\left(x+\frac{\pi}{4}\right)\right)}{n}\right)\,\mathrm dx \\ &= -\frac{\pi\ln(2)}{4}-\sum_{n=1}^\infty\frac{1}{n}\left[\frac{\sin\left(\frac{\pi n}{2}\right)\cos(2nx)}{2n}+\frac{\cos\left(\frac{\pi n}{2}\right)\sin(2nx)}{2n}\right]\Bigg\vert_0^{\frac{\pi}{2}}\\ &= -\frac{\pi\ln(2)}{4}-\frac{1}{2}\sum_{n=1}^\infty\frac{\sin\left(\frac{2\pi n}{2}\right)-\sin\left(\frac{\pi n}{2}\right)}{n^2}\\ &=-\frac{\pi\ln(2)}{4}+\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\\ &= G-\frac{\pi\ln(2)}{4}.\end{align*}\]where we have recognised the series for Catalan’s constant in the second to last line.
Putting it all together yields
\[\int_0^{\frac{\pi}{2}}\frac{x\cdot\sin(x)}{\cos(x)+\sin(x)}\,\mathrm dx=\frac{\pi^2}{16}+\frac{G}{2}-\frac{\pi}{8}\ln(2).\]Problem 2
This problem is comparitively straightforward,
\[\begin{align*} \int_0^{\pi/2}x\ln(\tan(x))\,\mathrm dx &= \int_0^{\pi/2}x\ln(\sin(x))\,\mathrm dx-\int_0^{\pi/2}x\ln(\cos(x))\,\mathrm dx \\ &= \int_0^{\pi/2}x\left(-\ln(2)-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\right)\,\mathrm dx + \int_0^{\pi/2}x\left(\ln(2)+\sum_{n=1}^\infty(-1)^n\frac{\cos(2nx)}{n}\right)\,\mathrm dx \\ &= -2\sum_{k=0}^\infty\frac{1}{2k+1}\int_0^{\pi/2}x\cos(2(2k+1)x)\,\mathrm dx,\,\text{terms of even index cancel} \\ &= -2\sum_{k=0}^\infty\frac{1}{2k+1}\left[\frac{(2(2k+1))x\sin((2(2k+1))x)+\cos((2(2k+1))x)}{(2(2k+1))^2}\right]\Bigg\vert_0^{\pi/2},\\&\qquad \text{integrate by parts with }u=x\text{ and }\mathrm dv=\cos(2(2k+1)x)\,\mathrm dx \\&= -2\sum_{k=0}^\infty\frac{1}{2k+1}\cdot\frac{-2}{4(2k+1)^2} \\&= \sum_{k=0}^\infty\frac{1}{(2k+1)^3} \\&= \sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3} \\&= \zeta(3)-\frac{1}{8}\zeta(3) \\&= \frac{7}{8}\zeta(3).\end{align*}\]