Monthly Archives: September 2013

Deceptively tricky integral

Someone wanted us to evaluate the following integral: $$\int\limits_0^\pi dx{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}$$ This looks straightforward, although the factor of $x^2$ does seem to complicate things a little. Actually, a lot. The usual tricks to deal with powers leads us down a road that leads to interesting territory involving dilogarithms. Let’s define $$J(a) = \int_{-\pi}^{\pi} dx […]

Unlikely application of Parseval’s equality

A poster somehow got a numerical value for an integral and wondered if it could be expressed in terms of simple constants. Let $\operatorname{erfi}(x)$ be the imaginary error function $$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$ Consider the integral $$I=\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx.$$ Its numeric value is approximately $0.625773669454426$ The question is: Is it possible to express $I$ in a closed form […]

A nifty integral worked out via contour integration

Problem: evaluate the following integral: $$\int_0^{\infty}dx \frac{e^{-a x^2(x^2-\pi^2)}\cos(2\pi a x^3)}{\cosh x} $$ To evaluate this integral, consider the following integral in the complex plane: $$\oint_C \frac{dz}{\sinh{z}} e^{-a (z^2-\pi^2/4)^2}$$ where $C=C_1+C_2+C_3+C_4+C_5+C_6$ as illustrated below: $$\int_{C_1} \frac{dz}{\sinh{z}} e^{-a (z^2-\pi^2/4)^2} = \int_{i \pi/2}^{R+i \pi/2} \frac{dx}{\sinh{x}} e^{-a (x^2-\pi^2/4)^2}$$ $$\int_{C_2} \frac{dz}{\sinh{z}} e^{-a (z^2-\pi^2/4)^2} = i\int_{\pi/2}^{-\pi/2} \frac{dy}{\sinh{(R+iy)}} e^{-a [(R+i y)^2-\pi^2/4]^2} $$ […]