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Thread: Improper Integral

  1. #1
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    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.
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  2. #2
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    Re: Improper Integral

    Quote Originally Posted by TenaciousE View Post
    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 $\displaystyle \ln(39)$
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