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

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


Thanks in advance.

  • $\begingroup$ This does it in Mathematica, but it ain't pretty: Integrate[Cos[a*x]/(1 + x^2), {x, b, Infinity}, Assumptions -> {b > 0, a > 0}] // FullSimplify $\endgroup$ – user14717 Apr 25 '12 at 0:33
  • $\begingroup$ I'm thinking curiosity's going to win and I'm going to end up looking in a table of integrals to see if this is in there. $\endgroup$ – Nicholas Stull Apr 25 '12 at 0:43
  • $\begingroup$ Thanks for trying everyone. $\endgroup$ – Stuck_pls_help Apr 25 '12 at 0:45

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).

  • $\begingroup$ There may not be an elementary antiderivative for this particular function (I haven't worked on it, so I can't say for sure), but it is still possible to have a closed form solution to this integral. $\endgroup$ – Nicholas Stull Apr 25 '12 at 0:38
  • $\begingroup$ But you could differentiate any closed form solution to this integral with respect to b, so a closed form solution is the same as (the negative of) an elementary antiderivative. $\endgroup$ – user29743 Apr 25 '12 at 1:52
  • $\begingroup$ 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? $\endgroup$ – Nicholas Stull Apr 25 '12 at 2:51

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