## Monthly Archives: January 2014

### 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}} = […]