Need hints on how to integrate this

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Hello,

I need to evaluate the following integral with respect to x. If you guys can give me some hints on how to start off, it would be greatly appreciated.

f{x}(x) = \int\frac{\alpha\beta}{4}e^{(-\alpha|x| - \beta|y|)}dy

The integral is evaluate from -infinity to infinity

From this I know that \frac{\alpha\beta}{4} is a constant and can be taken out. I get confused on how to integrate the absolute value portion of it.

Thanks!
 
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I did some manipulation and am at the following part:

\frac{\alpha\beta}{4}\int e^{-\alpha|x|} e^{-\beta|y|} dy

=\frac{\alpha\beta}{4}e^{-\alpha|x|}\int e^{-\beta|y|} dy
evaluate from -infinity to infinity
 
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To where you are now, evaluating the integral should be easy, noting that the remaining integrand is an even function.
 
Figured it out. Thanks!
 
=\frac{\alpha\beta}{4}e^{-\alpha|x|}\int e^{-\beta|y|} dy
evaluate from -infinity to infinity

Since this is an even function, we can just multiply the function by 2 and take out the absolute value because of symmetry and taking the limits of integration accordingly.

Therefore,
\frac{\alpha\beta}{4}e^{-\alpha|x|} * 2 \int e^{-\beta y} dy
I have attached what this graph would look like as a function of x. There is symmetry about the origin and the graph is even, so this function is even. We can just multiply the function by 2 and evaluate the integral from 0 to infinity because of the symmetry.

=\frac{\alpha\beta}{2}e^{-\alpha|x|} \int e^{-\beta y} dy, evalualted from 0 to infinity.

This integral evaluates to:
\frac{\alpha\beta}{2}e^{-\alpha|x|} * \left(\frac{-1}{\beta}e^{-\beta y}\right) evaluated from 0 to infinity.

e^{-\infty} = 0
e^{0} = 1

\frac{\alpha\beta}{2}e^{-\alpha|x|} * (-1)(0-1)
=\frac{\alpha\beta}{2}e^{-\alpha|x|}

Hopefully I did this right! :smile:
 

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You know you can make LaTeX show limits of integration: \int_{-\infty}^{\infty}f(x)\,dx, etc.
 
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