Evaluate $\displaystyle \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx \ \ p,q>0$ .

Then argue that $\displaystyle \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int^{\infty}_{0} \frac{\cos (x^{p}) - \cos(x^{q})}{x} \ dx$ .

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- Jun 30th 2011, 02:37 PMRandom Variableimproper integral #5
Evaluate $\displaystyle \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx \ \ p,q>0$ .

Then argue that $\displaystyle \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int^{\infty}_{0} \frac{\cos (x^{p}) - \cos(x^{q})}{x} \ dx$ . - Jun 30th 2011, 06:49 PMDrexel28Re: improper integral #5
- Jun 30th 2011, 07:40 PMRandom VariableRe: improper integral #5
I didn't even think there was a way to write it as a double integral that would make evaluating it easier.

I used another integral representation of $\displaystyle \gamma $, namely that $\displaystyle \int^{\infty}_{0} \Big(\frac{1}{1+x} - e^{-x} \Big) \ \frac{dx}{x} = -\psi_{0}(1) = \gamma $.

Can you also show that $\displaystyle \int^{\infty}_{0} \frac{ e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int_{0}^{\infty} \frac{\cos(x^{p}) - e^{-x^{q}}}{x} \ dx $ ? - Jul 1st 2011, 01:29 PMRandom VariableRe: improper integral #5
Drexel28

Could you explain your reasoning for the second part of the question in more detail? I feel really stupid because I don't understand where the integral of the form you posted even comes into play.