# I Can't find a solution to this integral

1. Mar 23, 2016

### MarkoA

Hi,

I try to solve this integral, but I failed... I have tried Euler's formula but it only got more and more complex. Can anyone help me?

$$\int_{0}^{T}\int_{0}^{L}\frac{e^{2i(\omega t+\alpha x)}}{\sqrt{1-a e^{2i(\omega t+\alpha x)}}}dx dt$$

2. Mar 23, 2016

### Ssnow

Hi, you must decide respect what variable you want to start ..., after I suggest a substitution of the exponential ...

3. Mar 23, 2016

### MarkoA

Thanks for your post. I start with dx. I can write

$$e^{2i(\omega t + \alpha x)} = \cos{(2\omega t + 2\alpha x)} + i \sin{(2\omega t + 2\alpha x)}$$
and then for the first term
$$\cos{(2\omega t + 2\alpha x)} = \cos{(2\omega t)}\cos{(2\alpha x)} - \sin{(2\omega t)}\sin{(2\alpha x)}$$
Thus, the first integral to solve would still be:
$$\cos{(2\omega t)} \int_{0}^{L} \frac{\cos{(2\alpha x)}}{\sqrt{1-a e^{2i (\omega t + \alpha x)}}}$$

I can of course use Euler's formula in the root as well, but it does not help yet.

Or I substitute with
$$k = 2 i (\alpha x + \omega t)$$
$$dx = \frac{1}{2i\alpha} dk$$
and come up with an integral like
$$\frac{1}{2i\alpha} \int \frac{e^k}{\sqrt{1-ae^k}} dk$$

Last edited: Mar 23, 2016
4. Mar 23, 2016

### Ssnow

exactly! Now substitute $1-ae^{k}=w$ so $e^{k}dk=-\frac{dw}{a}$ and the integral become elementary ...

5. Mar 30, 2016

Thanks!