# Elementary exponential integral

## Main Question or Discussion Point

I'm sure this integral is easy, but could someone perhaps show the working of:

$$\int$$ e$$^{i2t}$$ dt between t and 0.

I've tried it with trigonometric identities and keep getting lost!

Cheers!

Char. Limit
Gold Member
Try a u-substitution.

u=2it
du=2i dt

AlephZero
Homework Helper
Why do you need to do a substitution?

It is exactly the same as

$$\int_0^t e^{at}\,dt$$

except that $$a$$ is a complex number.

Char. Limit
Gold Member
Although strictly speaking, it's bad form to use the same variable in your bounds and your integrand.

Cheers guys.

Yea true I guess, perhaps it should have been tau as the differential.

This relates to the integral of exp[ iwt], between t and 0. This is then squared, so i was going to just integrate exp [ 2iwt].

The answer is 2(1 - coswt) / w ....i cant seem to get this at all. Any ideas?

Char. Limit
Gold Member
Is it the integral that's squared, or just the integrand?

HallsofIvy
Homework Helper
The integral
$$\int_0^t e^{-2i\tau}d\tau$$
is
$$-\fra{1}{2i}e^{-2it}= \frac{i}{2}e^{-2it}$$

NOT
$$\frac{2(1- cos(\omega t))}{\omega}$$
or even
[tex]\frac{2(1- cos(2t))}{2}= 1- cos(2t)[/itex]