# Homework Help: Evaluate INT (5+cosx)e^-x from 0 to infinity

1. Jan 18, 2010

### IntegrateMe

ok so i got (5+cos x)/ex and i compared that with 1/ex

(they're both going from 0 to infinity).

Turns out that the integral of 1/ex from 0 to infinity converges to 1. But i don't know how to prove that our original function converges as well (which is the answer). Anyone care to help?

2. Jan 18, 2010

### Mentallic

Split the integral into: $$5\int_0^\infty{e^{-x}}dx+\int_0^\infty{cosxe^{-x}}dx$$

and since you know the first one is equal to 5, just use integration by parts to find the second and then add 5 to it.

3. Jan 18, 2010

### Dick

You could also compare your original integral with 6/e^x. Which is larger?

4. Jan 18, 2010

### jgens

If you're just trying to prove that the integral converges, the fact that $\int_0^{\infty}\exp{(-x)}\mathrm{d}x$ converges is enough to prove that the second integral in Mentallic's post converges. For example, $|\cos{(x)}| \leq 1$ from which it follows that $\exp{(-x)}|\cos{(x)}| = |\exp{(-x)}\cos{(x)}| \leq \exp{(-x)}$ and then using this inequality we can reduce the convergence of the second integral to mere trivialities. However, if you actually need to evaluate the integral, Mentallic's method is definately the best way to go.

Edit: I would follow Dick's advice since his suggestion is considerably simpler than mine.

5. Jan 18, 2010

### Dick

Nah, it's not that much simpler. Mine only works because (5+cos(x)) is nonnegative. If it were (1+cos(x)) IntegrateMe would definitely want to use your comparison.

6. Jan 18, 2010

### neelakash

First break the integrand into twp parts as shown by Mentallic.You can reduce the pain of integration by writing cos(x) as Re[exp(ix)];next solve the integration,which is now trivial.Then,take the real part of the resulting indefinite integral and put the limits.The lower limit gives you finite value and the upper limit gives you zero for the e^-x factor.