Contour integration to find real integral

[bdforbes]

$$g(z) = \frac{e^{imz}}{z^2}$$

m is real, m > 0

Integrate g(z) around a suitable closed contour to find the principle value of

$$\int^{\infty}_{-\infty}\frac{e^{imx}}{x^2}dx$$


Obviously parts of the contour must lie on the negative and positive real axes. There's a double pole at z=0 so we need to indent around it. My first instinct is to integrate anti-clockwise around a large semi-circle C_R in the upper half plane, from -R to -r on the negative real axis, clockwise around a small semi-circle C_r, and from r to R. In the limit r to 0, R to infinity, the desired integral appears as part of the contour integration.
But the integration around C_r seems to diverge as r goes to zero. I can't think of any other way to approach this problem, can anyone help?

 

[mighty2000]

I would approach this problem by first understanding the properties of the function g(z) and its behavior along the real axis. From the given information, we know that m is a real number greater than 0, which means that the exponential term in g(z) will always have a positive argument. This suggests that g(z) will never have any singularities or poles along the real axis, except for the double pole at z=0.

Based on this, I would propose a different contour for integration. Instead of using a large semi-circle in the upper half plane, I would use a keyhole contour. This contour is composed of a large circle in the upper half plane, a small circle around the origin with radius r, and two straight lines connecting the two circles. The radius r can be chosen small enough so that it avoids the double pole at z=0.

Integrating along this contour, we can use the Cauchy's Residue Theorem to evaluate the integral. The keyhole contour encloses the double pole at z=0, and the only other singularity of g(z) is at infinity, which is outside of the contour. This means that the integral along the contour is equal to the sum of the residues at the poles inside the contour.

Using the Residue Theorem, we can calculate the residues at z=0 and z=infinity. The residue at z=0 is equal to the limit of (z-0)^2 * g(z) as z approaches 0, which is simply m. The residue at z=infinity is equal to the limit of z^2 * g(z) as z approaches infinity, which is 0.

Therefore, the integral along the keyhole contour is equal to 2πi * m, where i is the imaginary unit. This means that the desired integral is equal to 2πi * m. This is the principal value of the integral, as it is the value obtained by taking the limit of the contour as r approaches 0 and R approaches infinity.

In summary, as a scientist, I would approach this problem by first understanding the properties of the function and using the Residue Theorem to find the principal value of the integral. By carefully choosing a suitable contour, we can avoid the divergence issue and accurately calculate the desired integral.