Splitting an integral with an absolute value

Koshi
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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



\int\stackrel{\infty}{-\infty}(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

\int\stackrel{0}{-\infty}(x/x_0)e2x/x_0dx+\int\stackrel{\infty}{0}(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?
 
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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)?
 
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



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

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

\int\stackrel{0}{-\infty}(x/x_0)e2x/x_0dx+\int\stackrel{\infty}{0}(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 \infty 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 \infty 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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