Some older complex analysis textbooks state that $ \displaystyle \int_{0}^{\infty}e^{-x^{2}} \ dx$ can't be evaluated using contour integration.

But that's now known not to be true, which makes me wonder if you can ever definitively state that a particular real definite integral can't be evaluated using contour integration.

Edit: (t.b.) a famous instance of the above claim is in Watson, Complex Integration and Cauchy's theorem (1914), page 79:

Watson's claim

  • 1
    $\begingroup$ Could you add references for your statements? E.g., which older complex analysis textbooks do state this where? $\endgroup$ – user20266 Mar 31 '12 at 19:29
  • 2
    $\begingroup$ @Thomas: see edit. $\endgroup$ – t.b. Mar 31 '12 at 20:02
  • 4
    $\begingroup$ How about considering an integral on the real line that is so bizarre that it cannot be a profile of some holomorphic function? $\endgroup$ – Sangchul Lee Mar 31 '12 at 22:55
  • 3
    $\begingroup$ While I find this question very interesting, I have to note that the formulation in the quoted text is quite prudent. It does circumvent the statement that the example cannot be evaluated. $\endgroup$ – Phira Apr 26 '12 at 13:36
  • 1
    $\begingroup$ @Yrogirg: (late answer...) quite simply by taking as contour the first quarter of the circle : $\displaystyle \int_0^1 x^2\,dx+\int_0^{\frac {\pi}2} e^{2i\phi}ie^{i\phi}\,d\phi +\int_1^0 (ix)^2 i\,dx=0\ $ that becomes $\ \displaystyle (1+i)\int_0^1 x^2\,dx=\frac {1+i}3 $. $\endgroup$ – Raymond Manzoni Jul 6 '12 at 8:36

There are such functions. For example, anything with infinitely many discontinuities. Take the Dirichlet function as an example; it is Lebesgue integrable, but one could not integrate it using the method of residues, which requires that there are only finitely many poles of the function on the real line.


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.