Solving Complex Integrals: Can I Treat i as Any Other Constant?

[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.