# Complex integration

Logarythmic
How do I solve an integral of the type

$$\int f(v) e^{iavx} dv$$ ?

Can I just treat i as any other constant?

Nick89
I'm not entirely sure I am correct about this but it seems logical to expand the complex exponent and integrate it further from there.
$$e^{i \phi} = \cos (\phi) + i \sin (\phi)$$

At a guess I would say yes, $$i$$ is a constant... Just a logical guess though...

Logarythmic
$$\int f(v) e^{iavx} dv = \int f(v) \left( \cos{avx} + i \sin{avx} \right) dv =$$
$$= \int f(v) \cos{avx} dv + i \int f(v) \sin{avx} dv$$

Maybe?

Pere Callahan
Yes, you can treat $i$ as a constant.

Or you can use Euler's formula and write it as the sum of cos and sin, yes.

Nick89
Using Euler's formula doesn't get rid of the $$i$$ ofcourse...

Logarythmic, looks fine by me as long as you put $$avx$$ in brackets ;)

Logarythmic
Got it, so
$$w(x) = \int_{-u_0}^{u_0} i2 \pi v e^{i2 \pi vx} dv = \frac{1}{\pi x^2} \left[ 2 \pi u_0 x \cos{(2 \pi u_0 x)} - \sin{(2 \pi u_0 x)} \right]$$

Nick89
I got the same, so I guess it's correct.