3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.