## Category Archives: Uncategorized

### Logs on a dogbone

The problem is to evaluate the following integral: $$\int_0^1 dx \frac{\log^2{x}}{\sqrt{x-x^2}}$$ The integral is well-suited to real methods but the question of whether this integral was suited at all to complex methods has kept me up for…many nights. At first, I did get the result in terms of a limit of another integral. But […]

### Real integral evaluation via the residue theorem with two branch points and a log-squared term

Every so often there comes an integral that I see as a major teaching opportunity in complex integration applications. This integral represents one such opportunity. Problem: Evaluate the definite integral $$\int_0^{\infty} dx \, \frac{\log^2{x} \, \log{(1+x)}}{1+x^2}$$ This integral may be evaluated using the residue theorem. The analysis involves two branch points with respective branch […]

### A simple but nifty inequality

Problem: Given $f:[0,1] \to \mathbb{R}$ is integrable over $[0,1]$, and that $$\int_0^1 dx \, f(x) = \int_0^1 dx \, x f(x) = 1$$ show that $$\int_0^1 dx \, f(x)^2 \ge 4$$ The way to the solution here is not trivial. I started by always recognizing that, with integral inequalities, it never hurts to start with […]

### Another crazy integral, another clinic on using the Residue Theorem

The problem posed here is to show that $$\int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} = \frac{4}{5}$$ The OP actually seemed to know what he was doing but could not get the correct residue that would allow him to get the stated result. In showing a link to Wolfram Alpha, the OP revealed that he […]

### Another integral that Mathematica cannot do

The integral to evaluate is $$\int_0^{\infty} dx \frac{\sin{\left (\pi x^2 \right )}}{\sinh^2{(\pi x)}} \cosh{(\pi x)}$$ Given the trig functions in the integrand, it makes sense to use the residue theorem based on a complex integral around a rectangular contour. As has been my experience with these integrals, the integrand of the complex integral will not […]

### Computing an integral over an absolute value using Cauchy’s theorem

The problem is to compute the following integral: $$\int_{-1}^1 dx \frac{|x-y|^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}$$ I will show how to compute this integral using Cauchy’s theorem. It was remarked that it should not be possible to use Cauchy’s theorem, as Cauchy’s theorem only applies to analytic functions, and an absolute value certainly does not qualify. True. Nevertheless, for the […]

### Integral with two branch cuts II

The problem here is to compute $$\int_0^\infty \log(1+tx)t^{-p-1}dt$$ where $p\in(0,1)$ and $x>0$. This is a great problem for contour integration. Just tricky enough to be really interesting. What makes it interesting is that there are two functions in the integrand needing their own separate branch cuts. One must keep in mind that each function only […]

### Integral with two branch cuts

The problem is to evaluate the following integral: $$\int_{-1}^1 dx \frac{\log{(x+a)}}{(x+b) \sqrt{1-x^2}}$$ where $a \gt 1$ and $|b| \lt 1$. It should be obvious to those who spend time around these integrals that this integral does not converge as stated. However, we only have a simple pole at $x=-b$ so that we can compute […]