# How does the integral $\int_{D_C} e^{ia z}P(z)/Q(z)\,\mathrm{d}z$ blow up.

In my book I have a theorem that goes something like the following

Let $P(x)$ be $Q(x)$ polynomials such that $\deg(Q) \geq \deg(P) + 2$. Then \begin{align*} \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} e^{iax} \,\mathrm{d}x = 2 \pi i \frac{a}{|a|} \sum_{k=1}^{m} \mathrm{Res}\left[ \frac{P(x)}{Q(x)} , z_k \right] \end{align*} where $a$ is a real constant and $z_1,\,\ldots\,,z_m$ are the singularities to $P(x)/Q(x)$ in the upper halfplane if $a>0$ and the lower halfplane if $a<0$.

I am having a bit of problems understanding this. I know the reason why we have to switch contour is that $e^{iz}$ blows up. But I really can not see why or how it blows up.

Take the canonical example for why this theorem is useful $$J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1}$$

Now the clue here is to show that the integral along the curve tends to zero, and then use the residue theorem. To show that the integral tends to zero I did this \begin{align*} \left| \int_{C_1} \frac{e^{iz}}{1+z^2} \,\mathrm{d}z\right| \leq \int_{C_1} \left| \frac{e^{iz}}{1+z^2} \right| \,\mathrm{d}z \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{\left|1+z^2\right|} \right) \int_{C_1} |e^{iz}| \,\mathrm{d}z \end{align*} where the $ML$-inequality was used in the last inequality On the circle with radius $R$ we have $|e^{iz}|=R$, and we can use the inequality $|a+b|\leq|a|-|b|$ to simplify further \begin{align*} \left|\int_{D_R} \frac{e^{iz}}{1+z^2}\,\mathrm{d}z \right| \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{|z|^2-1}\right) \int_{D_R} R \,\mathrm{d}z \leq \pi \frac{R}{R^2-1} \end{align*} which tends to zero as $R \to \infty$ as wanted. but if one instead had $e^{-iz}$ then the theorem states that one has to use the lower half plane. But I do not see where my calculations err if one persists in using the contour in the upper half plane? What goes wrong, and why does it go wrong? $|e^{-iz}|$ should still be $R$ on the semi-circle.

How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane?

How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane?

The integral does - in the given situation - not diverge. If we let

$$I(a) = \int_{-\infty}^\infty \frac{P(x)}{Q(x)}e^{iax}\,dx,$$

which due to the assumption on the degrees exists as a Lebesgue integral as well as as an improper Riemann integral if $Q$ has no zeros on the real line, for large enough $R$, we have

$$-2\pi i \sum_{\operatorname{Im} \zeta < 0} \operatorname{Res} \left(\frac{P(z)}{Q(z)}e^{iaz};\zeta\right) = \int_{-R}^R \frac{P(x)}{Q(x)}e^{iax}\,dx - \int_{C_R} \frac{P(z)}{Q(z)}e^{iaz}\,dz,$$

where $C_R$ is the semicircle with centre $0$ and radius $R$ in the lower half-plane, traversed from $-R$ to $R$ (due to the choice of sign), by the residue theorem. Rearranging, we obtain

$$\lim_{R\to\infty} \int_{C_R} \frac{P(z)}{Q(z)}e^{iaz}\,dz = I(a) + 2\pi i \sum_{\operatorname{Im} \zeta < 0} \operatorname{Res} \left(\frac{P(z)}{Q(z)}e^{iaz};\zeta\right),$$

so it converges to a finite limit also for $a > 0$. However, that limit is in general not $0$:

For the example we find - with $a > 0$ -

$$I(a) = 2\pi i \operatorname{Res}\left(\frac{e^{iaz}}{z^2+1}; i\right) = 2\pi i \frac{e^{-a}}{2i} = \frac{\pi}{e^a}$$

using the semicircle in the upper half-plane, and thus

$$\lim_{R\to\infty} \int_{C_R} \frac{P(z)}{Q(z)}e^{iaz}\,dz = \frac{\pi}{e^a} + 2\pi i \operatorname{Res}\left(\frac{e^{iaz}}{z^2+1}; -i\right) = \frac{\pi}{ e^a} +2\pi i \frac{e^a}{-2i} = -2\pi\sinh a.$$

The point is: Even if the limit happens to be $0$, you cannot use it to determine $I(a)$ unless you can determine the limit in some way.

By choosing the semicircle in the correct half-plane, you get an estimate of the integrand that yields

$$\lim_{R\to\infty} \int_{C_R} \frac{P(z)}{Q(z)}e^{iaz}\,dz = 0$$

since $\lvert e^{iaz}\rvert = e^{-a\operatorname{Im} z}$ is bounded in the half-plane you choose, giving an $O(R^{-2})$ estimate of the integrand, and an $O(R^{-1})$ estimate for the integral by the standard estimate ($ML$-inequality). In the other half-plane, $e^{iaz}$ grows exponentially for $\lvert \operatorname{Im} z\rvert \to +\infty$, and thus you cannot establish the existence or value of the limit by using estimates, since there you have

$$\lim_{R\to\infty} \int_{C_R} \left\lvert \frac{P(z)}{Q(z)}e^{iaz}\right\rvert\,\lvert dz\rvert = +\infty,$$

the rational function being estimated below by $\frac{c}{\lvert z\rvert^2}$, and the exponential factor having modulus $e^{\lvert a\operatorname{Im} z\rvert}$.

Write $z=R e^{i \theta}$ for $\theta \in [0,\pi]$. Then

$$e^{i z} = e^{i R \cos{\theta}} \, e^{-R \sin{\theta}}$$

The magnitude of the integral over the arc is then bounded by

$$\frac{R}{R^2-1} \int_0^{\pi} d\theta \, e^{-R \sin{\theta}} \le \frac{2R}{R^2-1} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{\pi}{R^2-1}$$

Meanwhile

$$e^{-i z} = e^{-i R \cos{\theta}} \, e^{R \sin{\theta}}$$

so that the corresponding integral blows up as $R \to \infty$.

• say you pick the bottom semi-circle. You would then get $\int_0^\pi e^{R \sin \theta} \,\mathrm{d}\theta$ which tends to infinity so $|\int_{C_R}|\leq \infty$ which is not very useful. Can one say anything more about this integral? Like if it is finite or infinite, and if it is greater than zero? – N3buchadnezzar Apr 3 '14 at 13:49
• If you choose the botom semicircle, then the integral involving $e^{i z}$ blows up, and that involving $e^{-i z}$ vanishes, just as demonstrated above. – Ron Gordon Apr 3 '14 at 13:50
• Okay, so if we use the lower half plane we obtain the following "useful" estimate for the integral around the semi-circle $$\lim_{R\to \infty}\left| \int_{C_R}\mathrm{d}x \frac{e^{ix}}{1+x^2}\right| \leq \lim_{R\to \infty}\frac{R}{R^2-1} \int_0^{\pi} d\theta \, e^{R \sin{\theta}} \leq \infty$$ So I see now that this is not a useful estimate at all, since the integral does not tend to zero. Is there any way to say that the curve integral is finite or greater than zero, without calculating $I(a)$? – N3buchadnezzar Apr 4 '14 at 8:45
• If the integral involving $e^{iz}$ blew up along bottom half of the semicircle, that would mean that $\lim_{R \to \infty} \int_{-R}^{R} \frac{e^{ix}}{1+x^{2}} \ dx$ is not finite, which would imply that $\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^{2}} \ dx$ doesn't converge. This is essentially what Daniel Fischer is saying in the other post. And numerical calculations suggest that the integral is indeed approaching $2 \pi \sinh(1)$ if the semicircle is traversed clockwise. – Random Variable Apr 5 '14 at 15:41