By considering the contour integral aroun a 'D-contour' of the function $$f(z) = \frac{ze^{iaz}}{z^2 + 6 + 25}$$ for $a>0$, show that

$$\int\frac{x\sin(ax)}{x^2 +6x +25}dx = \frac{\pi}{4}e^{-4a}\left(4\cos(3a) + 3\sin(3a)\right).$$

I have the poles $-3 \pm 4i$ but the pole $-3 - 4i$ lies outside of my contour.

So the integral around the contour is $$\int f(z)dz = 2\pi i\frac{(-3 + 4i)e^{ia(-3+4i)}}{2(-3+4i)+6}= \frac{\pi}{4}e^{-4a}(-3+4i)e^{-3ia}$$

How do I show that the integral around the contour is equal to $$\int^\infty_{-\infty} \frac{xe^{iax}}{x^2 + 6 + 25}?$$

And I can therefore find the integral of the imaginary part and gain the required result. I nearly have the answer I'm just missing that step


The integral about the semicircular arc of radius $R$ is

$$i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} e^{i a R e^{i \theta}}}{R^2 e^{i 2 \theta} + 6 R e^{i \theta} + 25} $$

This has a magnitude bounded by

$$\frac{R^2}{R^2-6 R+25} \int_0^{\pi} d\theta \, e^{-a R \sin{\theta}} \le \frac{R^2}{R^2-6 R+25} \int_0^{\pi} d\theta \, e^{-2 a R \theta/\pi} \le \frac{\pi}{2 a} \frac{R}{R^2-6 R+25}$$

which vanishes as $R \to \infty$.

Thus the contour integral is equal to the integral over the real line.

  • $\begingroup$ Please can you explain why it vanishes? I was told that this would give O(1) as R tends to infinity $\endgroup$ – Nique Jan 19 '16 at 21:00
  • 2
    $\begingroup$ @Nique: This is an example of Jordan's Lemma - when you have an oscillating function in the integrand, the rest of the integrand need only vanish faster than $O(1)$ for the integral to converge. You can see that the exponential contributes a factor of $1/R$ to the magnitude here. $\endgroup$ – Ron Gordon Jan 19 '16 at 21:02

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