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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 $$ J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1} $$

Contour

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?

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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 $e^{\lvert a\operatorname{Im} z\rvert}$.

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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 $R \to \infty$.

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  • $\begingroup$ 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? $\endgroup$ – N3buchadnezzar Apr 3 '14 at 13:49
  • $\begingroup$ 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. $\endgroup$ – Ron Gordon Apr 3 '14 at 13:50
  • $\begingroup$ 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 $I(a)$? $\endgroup$ – N3buchadnezzar Apr 4 '14 at 8:45
  • $\begingroup$ 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. $\endgroup$ – Random Variable Apr 5 '14 at 15:41

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