Integration by Parts separately

1. Jan 19, 2010

cse63146

1. The problem statement, all variables and given/known data

Integrate: $$-\frac{2}{\theta} \int^{\infty}_0 y e^{-2y/\theta} dy + \frac{2}{\theta} \int^{\infty}_0 y e^{-y/\theta}dy$$

2. Relevant equations

3. The attempt at a solution

Let u = y/theta; y=u*theta; dy = du*theta, which becomes

$$-2 \int^{\infty}_0 u \theta e^{-2u} du + 2\int^{\infty}_0 u \theta e^{-u}du$$

Doing each integral seperately and then adding them up:

$$-2 \int^{\infty}_0 u \theta e^{-2u} du = u\theta e^{-2u} |^{\infty}_0 - 2 \theta \int^{\infty}_0 e^{-2u} du = \theta e^{-2u} |^{\infty}_0 = - \theta$$

$$2\int^{\infty}_0 u \theta e^{-u}du = -2u \theta e^{-u}|^{\infty}_0 + 2 \theta \int^{\infty}_0 e^{-u} du =-2 \theta e^{-u} |^{\infty}_0 = 2 \theta$$

When I add them up, I get theta, but the answer is supposed to be (3/2)theta. Where did I make the mistake?

2. Jan 19, 2010

Dick

Well, you just dropped a '2' factor. Maybe more than once. For example, why doesn't your boundary term in the first integration by parts have a '2' in it? I really don't want to check every term. But you should have gotten -theta/2 for the first integral.

3. Jan 19, 2010

cse63146

I did forget a 2 in the first integration by parts, but it equals to 0 so it wouldn't make much of a difference. But I did realise my mistake. I didnt diffrentiate the e^-2u properly in the second part.

4. Jan 19, 2010

Dick

That's true. Thanks for helping with the checking work!