# Integration by Parts

• Feb 6th 2013, 10:33 PM
SC313
Integration by Parts
Hi, I am trying to integrate this function, but do not know how to go past

$\int_{\theta}^{\infty}xe^{-(x-\theta)}dx$

Let $u = x, dv=e^{-(x-\theta)}, du = dx, v=-e^{-(x-\theta)}$

Therefore,

$-xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}$
$[\underbrace{-xe^{-\infty}}_{0}+xe^{-(\theta-\theta)}]-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]$

The X will not cancel, and I know that the answer, through Maple, is $1+\theta$.

The command I used in maple is
Code:

int(x*exp(-(x-theta)),x=theta..infinity);
. Perhaps I have done this incorrect.

Thank you!
• Feb 6th 2013, 11:43 PM
earboth
Re: Integration by Parts
Quote:

Originally Posted by SC313
Hi, I am trying to integrate this function, but do not know how to go past

$\int_{\theta}^{\infty}xe^{-(x-\theta)}dx$

...

Re-write the integral to:

$\int_{\theta}^{\infty}xe^{-(x-\theta)}dx = e^{\theta} \cdot \int_{\theta}^{\infty}xe^{-x}dx$

and then proceed.
• Feb 7th 2013, 01:39 AM
earboth
Re: Integration by Parts
Quote:

Originally Posted by SC313
...
Therefore,

$-xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}$
$[\underbrace{-xe^{-\infty}}_{0}+xe^{-(\theta-\theta)}]-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]$

The X will not cancel, and I know that the answer, through Maple, is $1+\theta$.

The command I used in maple is
Code:

int(x*exp(-(x-theta)),x=theta..infinity);
. Perhaps I have done this incorrect.

Thank you!

You stopped exactly one step befor the finish:

$-xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}$
$[\underbrace{-xe^{-\infty}}_{0}+\underbrace{xe^{-(\theta-\theta)}}_{=x}]\underbrace{-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]}_{-(0-1)}$

Expand the last bracket and you'll get $x + 1$ as the final result.
• Feb 7th 2013, 01:48 AM
Prove It
Re: Integration by Parts
The OP seems to have forgotten that when evaluating at the lower limit of \displaystyle \begin{align*} \theta \end{align*} that ALL values of x are replaced with \displaystyle \begin{align*} \theta \end{align*}.

Also for the upper limit of \displaystyle \begin{align*} \infty \end{align*} it should really be

\displaystyle \begin{align*} \lim_{\epsilon \to \infty} -\epsilon \, e^{-\left( \epsilon - \theta \right) } &= \lim_{\epsilon \to \infty} -\frac{\epsilon}{e^{\epsilon - \theta}} \\ &= \lim_{\epsilon \to \infty} -\frac{1}{e^{\epsilon - \theta}} \textrm{ by L'Hospital's Rule} \\ &= 0 \end{align*}
• Feb 7th 2013, 05:18 AM
earboth
Re: Integration by Parts
Quote:

Originally Posted by Prove It
The OP seems to have forgotten that when evaluating at the lower limit of \displaystyle \begin{align*} \theta \end{align*} that ALL values of x are replaced with \displaystyle \begin{align*} \theta \end{align*}.

...

Not only the OP, the same happened to me too (Worried)
• Feb 7th 2013, 07:12 AM
MINOANMAN
Re: Integration by Parts
The final answer is zero for this improper integral.
Minoas