# The Limit of a Complex Integral

1. Mar 20, 2013

### Bachelier

Though it is not homework I posted this here, hopefully it'll get more action. Thanks

given $$\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w$$ where $w \in \mathbb{C}$ with $Im(w) \geqslant 0$ and $|w| = \xi$

want to evaluate the behavior of the Integral as $\xi \rightarrow \infty$

So I have $$\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \int_{\Lambda} \left|\frac{e^{iw}}{w^2 + 1} \right| \mathrm{d}w$$

we have $|exp(iw)| \leqslant 1$

so we're left with the denominator. We can get:

$\left| \frac{1}{w^2 + 1} \right| \leqslant \frac{1}{\xi^2 - i^2}$

Hence the integral is:

$$\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \frac{ \pi \xi}{\xi^2 + 1}$$

Thus it goes to $0$ if $\xi$ goes to $\infty$

Last edited: Mar 20, 2013
2. Mar 20, 2013

### Jorriss

What are you integrating over? By saying |w| > 0 IMw > 0 are you indicating we are integrating over a semicircle?

3. Mar 20, 2013

### Bachelier

Exactly, my region $\Lambda$ is where the Imaginary part of $w$ is positive. So we are in the upper half-circle.

4. Mar 20, 2013

### Bachelier

And I am going to let $\xi$ explode, to use a Physicist term.

5. Mar 20, 2013

### Jorriss

Nevermind. You're not enclosing the real axis.

6. Mar 20, 2013

### jbunniii

Your estimate and conclusion look fine to me.