2
$\begingroup$

I want to find $$ \int_0^\infty \frac{1-\cos(x)}{x^2}dx $$ The integrand is continuous at $0$, so $f(z):=\frac{1-\cos(z)}{z^2}$ is entire. By the Residue Theorem, $$ 0=\int_{C_R} f(z)dz+\int_{-R}^R f(z)dz, $$ where $C_R$ is the semicircular contour of radius $R$ centered at $0$ in the upper half-plane (oriented counterclockwise). Now $$ \int_{C_R}f(z)dz= \int_0^\pi \frac{1-\cos(Re^{i\theta})}{R^2e^{i2\theta}}Rie^{i\theta}d\theta=\int_0^\pi \frac{1-\cos(Re^{i\theta})}{R}ie^{-i\theta}d\theta, $$ which does not seem easily manageable.

Differentiating under the integral sign works naively, by letting $$ I(a):=\int_0^\infty \frac{1-\cos(ax)}{x^2}dx $$ then $I'(a)=\frac{\pi}{2}$ and $I(0)=0$. The issue is that the derivative of the integrand is not integrable, so the passage of the limit into the integral is not legitimate.

$\endgroup$
2
$\begingroup$

View your integrand as the real part of

$$\frac{1-e^{iz}}{z^2}$$

then integrate over an indented semicircle in the upper half-plane.

The details in full are worked on on page 44 of Stein and Shakarchi's "Complex Analysis"


I hope this helps ^_^

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.