2
$\begingroup$

Consider the following improper integral:

\begin{equation} \int_0^\infty \frac{\cos{2x}-1}{x^2}\;dx \end{equation}

I would like to evaluate it via contour integration (the path is a semicircle in the upper plane), but i have some problems: first, the only singularity would be $z=0$, but it is only an apparent singularity so the residue is $0$. There are no other singularity of interest, so the integral should be zero... But it can't be zero, so?

  • 0
    @GEdgar: oh yes, thanks for pointing that out.2012-02-19

2 Answers 2

1

$ \int_0^\infty\frac{\cos 2\,x-1}{x^2}\,dx=\frac12\,\int_{-\infty}^\infty\frac{\cos 2\,x-1}{x^2}\,dx=\frac12\,\int_{-\infty}^\infty\frac{\Re(e^{2ix}-1)}{x^2}\,dx. $ The key is the choice of function to integrate along a path. The function $ f(z)=\frac{e^{2iz}-1}{z^2}=\frac{2\,i}{z}-2-\frac{4\,i}{3}\,z+\cdots $ has a simple pole at $z=0$ with residue $2\,i$. Take $R>0$ large and $\epsilon>0$ small. Integrate along a path formed by the positively oriented semicircle of radius $R$ in the upper half plane ($C_R$), the interval $[-R,-\epsilon]$ ($C_\epsilon$), the semicircle of radius $\epsilon$ negatively oriented and the interval $[\epsilon,R]$ and take the limit as $R\to\infty$ and $\epsilon\to0$. The integral along the path is zero, $\lim_{R\to\infty}\int_{C_R}f(z)\,dz=0$, but $\lim_{R\to\infty}\int_{C_\epsilon}f(z)\,dz=?$.

  • 0
    $\lim_{R\to\infty}\int_{C_R}\frac{\cos 2\,z-1}{z^2}\,dz=\ne0.$2012-02-19
0

It is much easier to use Laplace Transform to calculate this improper integral. Recall that if $F(s)$ is the Laplace Transform of $f(x)$, then $\mathcal{L}\big\{\frac{f(x)}{x}\big\}=\int_s^\infty F(u)du.$ Let $f(x)=\cos 2x-1$; then $F(s)=\frac{s}{s^2+4}-\frac{1}{s}$. Thus \begin{eqnarray*} \mathcal{L}\big\{\frac{\cos 2x-1}{x}\big\}&=&\int_s^\infty\left(\frac{u}{u^2+4}-\frac{1}{u}\right)du\\ &=&\ln s-\frac{1}{2}\ln(s^2+4). \end{eqnarray*} Therefore \begin{eqnarray*} \mathcal{L}\big\{\frac{\cos 2x-1}{x^2}\big\}&=&\int_s^\infty\left(\ln u-\frac{1}{2}\ln(u^2+4)\right)du\\ &=&-\pi+2\arctan\frac{s}{2}-s\ln s+\frac{1}{2}\ln(s^2+4). \end{eqnarray*} So $\int_0^\infty\frac{\cos 2x-1}{x^2}dx=\lim_{s\to o^+}\left(-\pi+2\arctan\frac{s}{2}-s\ln s+\frac{1}{2}\ln(s^2+4)\right)=-\pi.$