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I'm very rusty with calculus, and I was hoping someone would be willing to help me with the following definite integral:

$\int_b^{\infty} \frac{\cos(ax)}{1+x^2} dx$

$b>0$

Thanks in advance.

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    Thanks for trying everyone.2012-04-25

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For $b = 0$ or $b = -\infty$ you can use contour integration. For other $b$ (assuming $a \neq 0$), you will have to do it numerically.

(If there were a formula in terms of $b$, you would have an elementary formula for an antiderivative of the integrand (just differentiate your formula with respect to $b$ and you'll recover the negative of the integrand!), but there is no elementary formula for an antiderivative of this integrand, I believe).

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    I had a counterexample in mind when I made that comment. If we evaluate $\int_0^b \sin(\frac{1}{x})\, dx$, we can differentiate the closed form solution (which is expressible in terms of the cosine integral) with respect to $b$, but that integrand does not necessarily admit an elementary antiderivative. Or am I missing something?2012-04-25