Recent content by polygamma

Hint:

Show that the residue of $\cot^{n}(z)$ at $z=0$ is $\sin \left( \frac{n \pi}{2}\right)$, $n \in \mathbb{N}$.

Homework Statement If \int_{0}^{1} f(x) g(x) \ dx converges, and assuming g(x) can be expanded in a Taylor series at x=0 that converges to g(x) for |x| < 1 (and perhaps for x= -1 as well), will it always be true that \int_{0}^{1} f(x) g(x) \ dx = \int_{0}^{1} f(x) \sum_{n=0}^{\infty}...

By integrating $ \displaystyle f(z) = \frac{e^{-z^{2}}}{z}$ around the appropriate contour, or otherwise, show that for $a>0$, $$ \int_{0}^{\infty} e^{-x^{2}} \ \frac{a \cos (2ax) + x \sin(2ax)}{x^{2}+a^{2}} \ dx = \frac{\pi}{2}e^{-a^{2}}.$$

Hint:

Show that for $|a| > |b| $, $$\int_{0}^{\infty} \frac{\sinh bx}{\cosh ax + \cosh bx} \ dx = 2 \ln 2 \ \frac{b}{a^{2}-b^{2}} .$$

There is a quasi-reflection formula for the trilogarithm, but it also involves $\text{Li}_{3} \left( 1-\frac{1}{x} \right)$. But I'm sure you already knew that.

You could make the substitution $u = x^{n}$ and then justify bringing the limit inside of the integral. Doing so you'll see that the answer is $\frac{1}{2}$.

I'm assuming that $a$ and $b$ are positive parameters. $$ \frac{2}{i} \int_{|z|=1} \frac{1}{bz^{2}+2az+b} \ dz = \frac{2}{i} \int_{|z|=1} \frac{1}{b(z+\frac{a}{b} - \frac{\sqrt{a^{2}-b^{2}}}{b})(z+ \frac{a}{b} + \frac{\sqrt{a^{2}-b^{2}}}{b})} $$ $$ = \frac{2}{ib} \int_{|z|=1}...

That combination of function and contour won't work. The log term will vanish. You either need to consider $ \displaystyle f(z) = \frac{\log^{2} z}{1+z^{2}} $, or you could move the branch cut from the negative real axis to the negative imaginary axis and integrate around a contour that...

Nice. It never crossed my mind to do that.