Splitting an integral with an absolute value

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 4K views
Koshi
Messages
18
Reaction score
0
This is a physics problem, but I only need help with the calculus portion of it. I was having trouble figuring out how to split the integral to properly integrate.


Homework Statement




Homework Equations



[tex]\int[/tex][tex]\stackrel{\infty}{-\infty}[/tex](x/x_0)e-2|x|/x_0dx

where x_o is a constant

The Attempt at a Solution



I was wondering how to write the lower part of the integral. What I have is

[tex]\int[/tex][tex]\stackrel{0}{-\infty}[/tex](x/x_0)e2x/x_0dx+[tex]\int[/tex][tex]\stackrel{\infty}{0}[/tex](x/x_0)e-2x/x_0dx

Is that right or should I keep the negative in front of the 2 in the lower integral?
 
Physics news on Phys.org
Fixed the LaTeX in your first integral. Double-click it to see what I did.
Here are some tips:
Use only a single pair of [ tex] and [ /tex] tags (without the leading spaces that I show) for a given expression. You had tex tags around almost every item.
Don't mix [ sup] and [ sub] tags inside [ tex] tags. They don't work. Instead use ^{} for superscripts and _{} for subscripts.
Koshi said:
This is a physics problem, but I only need help with the calculus portion of it. I was having trouble figuring out how to split the integral to properly integrate.


Homework Statement




Homework Equations



[tex]\int_{-\infty}^{\infty}(x/x_0)e^{-2|x|/x_0} dx[/tex]

where x_o is a constant

The Attempt at a Solution



I was wondering how to write the lower part of the integral. What I have is

[tex]\int[/tex][tex]\stackrel{0}{-\infty}[/tex](x/x_0)e2x/x_0dx+[tex]\int[/tex][tex]\stackrel{\infty}{0}[/tex](x/x_0)e-2x/x_0dx

Is that right or should I keep the negative in front of the 2 in the lower integral?
 
Mark44 said:
Fixed the LaTeX in your first integral. Double-click it to see what I did.

Thanks for the fix. I was struggling with it :P
 
Dick said:
You've got it right. But there is an easy way to do your integral. If f(x) is the integrand, how does f(x) compare with f(-x)?

I was thinking I could just double the integral from 0 to [tex]\infty[/tex] but because there was an x/x_0 in front and I would have to integrate by parts, I reasoned against it.

I wouldn't know how else to simplify it.
 
Koshi said:
I was thinking I could just double the integral from 0 to [tex]\infty[/tex] but because there was an x/x_0 in front and I would have to integrate by parts, I reasoned against it.

I wouldn't know how else to simplify it.

Yes, if you want to do the integral from 0 to infinity you should integrate by parts. But you can't double it to get the whole integral because the integral from -infinity to 0 isn't the same. How are the two integrals related? You could, of course, just do the integration and find out... But you would be going to a lot of extra work.