# Improper Integral

• May 1st 2012, 05:21 PM
TenaciousE
Improper Integral
The problem is a definite integral from 0 to infinity for the function f(x) = 38/(z^2 + 40z + 39).
The problem here is that I don't know how to solve the integral. I can break the denominator into (z+39)(z+1) but I don't know where to go from here.
• May 1st 2012, 05:32 PM
skeeter
Re: Improper Integral
Quote:

Originally Posted by TenaciousE
The problem is a definite integral from 0 to infinity for the function f(x) = 38/(z^2 + 40z + 39).
The problem here is that I don't know how to solve the integral. I can break the denominator into (z+39)(z+1) but I don't know where to go from here.

f(x) or f(z) ?

method of partial fractions ...

$\displaystyle \frac{38}{(z+39)(z+1)} = \frac{A}{z+39} + \frac{B}{z+1}$

$\displaystyle 38 = A(z+1) + B(z+39)$

let $\displaystyle z = -1$ ...

$\displaystyle 38 = B(38)$ ... $\displaystyle B = 1$

let $\displaystyle z = -39$ ...

$\displaystyle 38 = A(-38)$ ... $\displaystyle A = -1$

$\displaystyle \lim_{b \to \infty} \int_0^b \frac{1}{z+1} - \frac{1}{z+39} \, dx$

you should get the limit of $\displaystyle \ln(39)$