Thread: Residue Theorem Integrals (Complex Analysis)

1. Residue Theorem Integrals (Complex Analysis)

Compute these integrals:

(1) $\displaystyle \int_{-\infty}^{\infty}$ $\displaystyle \frac{x^2}{x^4-4x^2+5}dx$

(2) $\displaystyle \int_{-\infty}^{\infty}$ $\displaystyle \frac{cos{\alpha}x}{(x^2+1)(x^2+4)}dx$

(3) $\displaystyle \int_{-\infty}^{\infty}$ $\displaystyle \frac{xsinx}{x^4+1}dx$

The first problem deals with an integral of a rational function. The other two deal with integrals over the real axis involving trig functions.
I am suppose to do these similar to the examples in my book. However, it is still complicated for me and I don't understand.. Any help would be greatly appreciated, thanks!

2. They can all be done with straight residue integration. Just use an upper half-disc ($\displaystyle H_u$) right? They can be made to all go to zero along the upper half-arc contour if the trigs are expressed in exponential forms , and the straight contours across the real axis are the integrals you want. I'll do the third one:

\displaystyle \begin{aligned} \int_{-\infty}^{\infty}\frac{x\sin(x)}{x^4+1}dx &= \text{Im}\left\{\lim_{R\to\infty}\mathop\oint\limi ts_{H_u}e^{iz}\frac{z}{z^4+1}dz\right\} \\ &= 2\pi i\sum_{p}\mathop\text{Res}\left(\frac{ze^{iz}}{z^4 +1}\right);\quad p=\{e^{\pi i/4},e^{3\pi i/4}\} \end{aligned}

The integral is the imaginary component of the limit of a contour integral around the upper half-disc as the radius goes to infinity. This contour enclosed only two poles given by the set $\displaystyle p$. In the limit, the integral over the half-arc goes to zero leaving only the segment over the real axis remaining. Since $\displaystyle x\sin(x)=x\text{Im}(e^{ix})$, then the desired integral is the imaginary part of the contour integral which is simply $\displaystyle 2\pi i$ times the two residues.

Similar for the second one and the first one just use residue integration directly on it using the same contour.

3. The answer to the 3rd one is supposed to be:

$\displaystyle \pi$$\displaystyle exp(\frac{-\sqrt{2}}{2})sin(\frac{\sqrt{2}}{2}) How do I get it to be that answer? 4. Originally Posted by shadow_2145 The answer to the 3rd one is supposed to be: \displaystyle \pi$$\displaystyle exp(\frac{-\sqrt{2}}{2})sin(\frac{\sqrt{2}}{2})$

How do I get it to be that answer?
As shawsend told us the answer is:
$\displaystyle 2\pi i\sum_{p}\mathop\text{Res}\left(\frac{ze^{iz}}{z^4 +1}\right);\quad p=\{e^{\pi i/4},e^{3\pi i/4}\}$

Now the function $\displaystyle \frac{ze^{iz}}{z^4+1}$ is of the form $\displaystyle \frac{f(z)}{g(z)}$ and $\displaystyle g'(p)\not = 0$. Thus, its residue is given by $\displaystyle \frac{f(p)}{g'(p)}$.

(1) $\displaystyle \int_{-\infty}^{\infty}$ $\displaystyle \frac{x^2}{x^4-4x^2+5}dx$

the following file shows the answer

6. [quote=Zamorano;204189]
(1) $\displaystyle \int_{-\infty}^{\infty}$ $\displaystyle \frac{x^2}{x^4-4x^2+5}dx$