two wild and crazy integrals

[galactus]

Show that:

$$\int_{0}^{2\pi}\frac{(1-4\sin^{2}(x))\cos(2x)}{2-\cos(x)}dx=\frac{2\pi (91-52\sqrt{3})}{\sqrt{3}}$$


and

Evaluate:

$$\int_{0}^{\frac{\pi}{2}}\log\left(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x)\right)dx, \;\ 0<a<b$$

[galactus]

Sorry for the typo on the first integral. I do not know why I had an 'n' there. But, maybe we can generalize. I reckon these two are not interesting or too easy?.

But, for the case when n=1, a simple unit circle contour will suffice quite easily.

Use the usual $$\cos(x)=\frac{z+z^{-1}}{2}, \;\ \sin(x)=\frac{z-z^{-1}}{2i}, \;\ dx=\frac{1}{iz}dz$$

This results in

$$i\int_{|z|=1}\frac{(z^{4}+1)(z^{4}-z^{2}+1)}{z^{4}(z^{2}-4z+1)}dz$$

The poles inside the contour are $z=0, \;\ z=2-\sqrt{3}$

$$2\pi i Res\left(z=0\right)=-104\pi$$

$$2\pi i Res\left(2-\sqrt{3}\right)=\frac{182\pi}{\sqrt{3}}$$

sum the residues and factor:

$$\frac{182\pi}{\sqrt{3}}-104\pi = \frac{2\pi}{\sqrt{3}}\left(91-52\sqrt{3}\right)$$


The other integral with the log is not too bad using DUTIS. Would anyone like a stab at it using contours?.

[galactus]

Well, since no one bit, I reckon I will post a little something on the log integral.

$$\int_{0}^{\frac{\pi}{2}}\log(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x))dx$$

Let $u=\tan(x), \;\ dx=\frac{1}{u^{2}+1}du$

$$\int_{0}^{\infty}\frac{\log\left(\frac{a^{2}u^{2}+b^{2}}{u^{2}+1}\right)}{u^{2}+1}du$$

$$=\int_{0}^{\infty}\frac{\log(a^{2}u^{2}+b^{2})}{u^{2}+1}du-\int_{0}^{\infty}\frac{\log(u^{2}+1)}{u^{2}+1}du$$

The right integral is rather famous and evaluates to $\pi \log(2)$, so no need to step through it.

The left one write as:

$$\int_{0}^{\infty}\frac{\log(u^{2}+\frac{b^{2}}{a^{2}})}{u^{2}+1}du+2\log(a)\int_{0}^{\infty}\frac{1}{u^{2}+1}du$$

Let $c=b/a$, and thus far we have:

$$\int_{0}^{\infty}\frac{\log(u^{2}+c^{2})}{u^{2}+1}du-\pi \log(a)+\pi\log(2)$$

Now, for the remaining integral, consider $$f(z)=\frac{\log(c\cdot \sin\theta -i(z+c\cdot \cos\theta))}{z^{2}+1}$$.

There is a branch point at $z=-ce^{i\theta}$, so we can avoid it altogether and use the UHP.

The residue at $z=i$ is $2\pi i Res(f(z),i) = \pi \log(c+1)=\pi \log(a+b)-\pi \log(a)$

So,

$$\int_{0}^{\infty}\frac{\log(a^{2}u^{2}+b^{2})}{u^{2}+1}du-\pi \log(a)+\pi log(2)=\pi \log(a+b)-\pi \log(a)$$

which implies that:

$$\int_{0}^{\frac{\pi}{2}}\log(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x))dx=\pi \log\left(\frac{a+b}{2}\right)$$