# Tough integration

1. Sep 27, 2008

### rhuelu

integral of x*(1/2b)*exp(-abs(x-a)/b)

sorry about the format, I dont know how to use the signs.

this looks like an integration by parts, but I'm not really seeing how to work it out

thanks!

2. Sep 27, 2008

### arildno

Would that be something like:
$$\int{x}\frac{1}{2b}e^{-\frac{|x-a|}{b}}dx$$??

3. Sep 27, 2008

### rhuelu

Exactly. Thanks.

4. Sep 27, 2008

### arildno

Can you pinpoint the problem or problems you have in finding an anti-derivative to that function?

5. Sep 27, 2008

### arildno

A helpful advice is to look at the DEFINITION of the absolute value function:
|y|=y if y>=0 and |y|=-y if y<0

Thus, for x-a>=0 (i.e, x>=a), we may simplify:
$$\int{x}\frac{1}{2b}e^{-\frac{|x-a|}{b}}dx=\int{x}\frac{1}{2b}e^{-\frac{x-a}{b}}dx=\frac{1}{2}e^{\frac{a}{b}}\int\frac{x}{b}e^{-\frac{x}{b}}dx=\frac{b}{2}\int{u}e^{-u}du,u=\frac{x}{b}$$
This is fairly trivial to anti-differentiate.

Make similar simplifications for the case x-a<0.

6. Sep 27, 2008

### rhuelu

I'm not really sure how to deal with the absolute value

7. Sep 27, 2008

### rhuelu

oh ok thanks

8. Sep 27, 2008

### rhuelu

how would I got about finding the value of the integral from negative infinity to positive infinity if I have 2 different expressions depending on the value of x-a?

9. Sep 27, 2008

### rhuelu

nevermind, i got it

10. Sep 27, 2008

### JG89

I haven't dealt with improper integrals yet, but my guess would be to integrate it over two separate intervals. One from negative infinite to 0 and the other from 0 to positive infinite.