Consider the function


where $\phi$ is a $C^\infty$ function. I want to compute the integral


where $t$ is a parameter which we can consider positive. For that I know that

$$I=\lim_{R\to \infty}\int_{-R}^{R}g(z)dz,$$

so that we can use contour integration. The problem is that in this case the pole is along the path over which we need to integrate. The usual way I've seem to deal with this is to consider the following paths:

  1. The interval $[-R,-\eta]$ being $\eta > 0$.
  2. The semicircle $C_{\eta}$ of radius $\eta$ centered at the origin, so that we traverse it starting at $-\eta$ and going all the way to $\eta$.
  3. The interval $[\eta,R]$,
  4. The semicircle $C_R$ of radius $R$ centered at the origin, so that we complete the loop.

We then have that inside the enclosed area there are no singularities of $g$. The integral on this whole path then is zero:


If we then let $R\to \infty$ we can easily see that the integral over $C_R$ goes to zero if $t>0$ thus we get

$$\int_{-\infty}^{-\eta}g(z)dz+\int_{\eta}^\infty g(z)dz=-\int_{C_\eta}g(z)dz,$$

now we just need to make $\eta\to 0$ to get what we want. The problem lies in computing the integral over $C_\eta$. We can parametrize the path as $z = \eta e^{i\theta}$, but then

$$\int_{C_\eta}g(z)dz=\int_{\pi}^{0}ie^{i(\eta \cos \theta + i\eta \sin \theta)t}d\theta,$$

but now I have no idea of what I do with this integral.

My question is: is my approach to this integration correct? If so, how do I compute this last integral? If not, how should I compute $I$ using contour integration?

  • $\begingroup$ 1.) in ur last integral $g(\eta e^{i\Theta})$ is missing 2.) because $\eta$ is infinitely small u might use a zeroth order taylor expansion of the integrand can u conclude? $\endgroup$ – tired Mar 26 '16 at 14:37

Check the corollary to the lemma in the most upvoted answer here: Definite integral calculation with poles at 0 and $\pm i\sqrt{3}$

In your case, since

$$\left.\text{Res}\left(\frac{e^{izt}\phi(z)}z\right)\right|_{z=0}=\lim_{z\to 0} e^{izt}\phi(z)=\phi(0)$$

so that then



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