Complex Analysis - Integration

**moolimanj** · June 26th 2007, 04:59 AM

Use a theorem to show that:

$\int_{-\infty}^{\infty}3sin2t/t(t^2+9)dt = (\pi i/3)(1-1/e^6)$

**ThePerfectHacker** · June 26th 2007, 06:29 AM

How can the integral that you gave, which is real valued possibly have $i$ in it? Check your problem again.

**ThePerfectHacker** · July 4th 2007, 07:51 PM

I was moving threads and found this one.
Let me try to find this integral.

Here we need to use two theorems.

Michael Jordan's Lemma: Let $f(z)$ be a meromorphic function in the upper half-plane with finitely many poles: $b_1,...,b_n$ in upper half-plane. Given that $f(z)\to 0 \mbox{ as }|z| \to \infty$ in the upper-half plane. Then for each positive number $a$ we have:
$\int_{-\infty}^{\infty} e^{iax} f(x) dx = 2\pi i \sum_{k=1}^n \mbox{res}(g,b_k)$ where $g(z) = e^{iaz}f(z)$

Lemma: Let $f(z)$ has a simple pole at $c$ with residue $\rho$ . Let $\gamma$ be the curve $\gamma(t) = c + re^{it} \mbox{ for }a\leq t\leq b$ .
Then, $\lim_{r\to 0^+} \int_{\gamma}f(z) dz = i \rho (b-a)$

Now we can evaluate,
$\int_{-\infty}^{\infty}\frac{3\sin 2x}{x(x^2+9)} dx$

Consider the meromorphic function,
$f(z) = \frac{3}{z(z^2+9)}$
The bad thing is that we cannot use Jordan's Lemma because the poles of this function are $z=0,z=\pm 3i$ . We disregard $z=-3i$ because we never will be working in the lower half plane, however $z=0$ lies on the real axis! So the conditions of Jordan's Lemma are not fullfiled. In order to avoid that we will use the other lemma by drawing a small circular contor around is "bad" point. Look at picture below.

The bottom is divided into three parts: $\gamma_1$ which is the left-line segment moving from $-\infty$ to $-r$ . Also, $\gamma_2$ which is the semi-circular contour (notice, this is negatively oriented). Finally, $\gamma_3$ which is moving from $r$ to $\infty$ .

Now the only pole in upper half plane of $g(z) = \frac{3e^{2zi}}{z(z^2+9)}$ is $z=3i$ . And $\mbox{res}(g,3i)=\lim_{z\to 3i} (z-3i)\cdot \frac{3e^{2zi}}{(z-3i)(z+3i)} = \frac{3e^{-6}}{6i} = - \frac{1}{2}ie^{-6}$

Now by this generalization of Michael Jordan's Lemma we have (notice the negative):
$\int_{-\infty}^{-r}\frac{3e^{2xi}}{x(x^2+9)} dx - \int_{\gamma_2} \frac{3e^{2zi}}{z(z^2+9)} + \int_r^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = 2\pi i \left( -\frac{1}{2}ie^{-6} \right) = \pi e^{-6}$

Now if we take the limit at $r\to 0^+$ by that Lemma we have:
$\int_{-\infty}^{0^-} \frac{3e^{2xi}}{x(x^2+9)}dx - \mbox{res}(g,0)\cdot (\pi - 0)\cdot i + \int_{0^+}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = \pi e^{-6}$

Since $\mbox{res}(g,0) = \lim_{z\to 0} z\cdot \frac{e^{2zi}}{z(z^2+9)} = \frac{1}{9}$

We have,
$\int_{-\infty}^{0^-} \frac{3e^{2xi}}{x(x^2+9)} dx + \int_{0^+}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx = \pi e^{-6}+ \frac{1}{9}\pi i$

Thus,
$\int_{-\infty}^{\infty} \frac{3e^{2xi}}{x(x^2+9)} dx= \int_{-\infty}^{\infty} \frac{3\cos 2x}{x(x^2+9)} + i \int_{-\infty}^{\infty} \frac{3\sin 2x}{x(x^2+9)} dx = \pi e^{-6}+\frac{1}{9}\pi i$

Thus,
$\int_{-\infty}^{\infty} \frac{3\cos 2x}{x(x^2+9)} \ dx = \pi e^{-6}$
And,
$\int_{-\infty}^{\infty} \frac{3\sin 2x}{x(x^2+9)} \ dx = \frac{1}{9}\pi$

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