Monthly Archives: January 2014

An integral that illustrates the beauty of contour integration methods

I attack many of the integrals in this blog using contour integration methods, some of which are obvious and some of which take a little more imagination. Here is one that illustrates the beauty an power of such methods as much as any other integral here. The problem is to derive the integral representation $$\sin […]

Funky triple integral

Note: There was a sign error in the original solution, which I have since fixed. Problem: Evaluate $$ \int _0 ^{\infty}\int _0 ^{\infty}\int _0 ^{\infty} \log(x)\log(y)\log(z)\cos(x^2+y^2+z^2)dzdydx$$ Solution: One way to attack this is to exploit the symmetry of the integral. Start by expanding the cosine term into individual pieces, i.e., $$\begin{align}\cos{(x^2+y^2+z^2)} &= \cos{x^2} \cos{(y^2+z^2)} – […]

An odd-looking integral of a square root of trig functions

The integral to evaluate is $$\int_0^{\pi/3} \big((\sqrt{3}\cos x-\sin x)\sin x\big)^{1/2}\cos x \,dx $$ A cursory glance at this specimen leads to exasperation, as the $\pi/3$ in the limit seems arbitrary, and the integrand seems devoid of an antiderivative. It turns out, however, that one way to attack this integral is to transform it into a […]

Integral of function with deceptive triple pole

One integral posted came from Hermite’s integral representation of the Hurwitz zeta function. The integral is not the most difficult that I have ever evaluated, but is interesting from a pedagogical point of view. The problem is to evaluate $$\int_0^{\infty} dx \frac{x}{(e^x-1) (x^2+4 \pi^2)^2}$$ This integral may be done via the residue theorem, by considering […]

Deceptively Tricky Integral II

Evaluate the following integral $$\int_0^{\pi/2} dx \frac{x^2}{1+\cos^2 x}$$ This integral is deviously difficult. It may look like it has the solution I provided for this integral, but, as you will see, there is an additional wrinkle. As in the linked solution, I will express the integral in a form in which I may attack via […]

Improper integral of a high power of log

I feel like I’ve developed something new in evaluating this integral. Previously, when confronted with an integral of log to the n times a function, I considered the integral of log to the n+1 times the function over a keyhole contour. This gave me the original integral, but also all of the other integrals of […]

Two integrals, each easier than it looks

I start with an integral that stymied a bunch of people for the better part of an hour. The following solution had a lot of people slapping their heads. The problem is to evaluate, for any real $\alpha$, the following integral: $$\int_0^{\pi/2} \frac{dx}{1+\tan^{\alpha}{x}}$$ Solution: use the fact that $$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$ i.e., $$\frac1{1+\tan^{\alpha}{x}} = […]