Need hints on how to integrate this

[skybox]

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!

 

[skybox]

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

 

[Gib Z]

To where you are now, evaluating the integral should be easy, noting that the remaining integrand is an even function.

 

[skybox]

Figured it out. Thanks!

 

[skybox]

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

 

[uman]

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