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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

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    Could you add references for your statements? E.g., which older complex analysis textbooks do state this where?2012-03-31
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    @Thomas: see edit.2012-03-31
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    How about considering an integral on the real line that is so bizarre that it cannot be a profile of some holomorphic function?2012-03-31
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    @sos440 Would you give an example? I'd vote for that answer. Would something like $\int_{\mathbb{R}}e^{-1/x^2}\cdot e^{-x^2}\,dx$ work?2012-04-05
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    $\int_0^1|\sin(1/x)|dx$?2012-04-11
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    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.2012-04-26
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    Huh, how do we compute $\int_0^1 x^2 dx$ with contour integration?2012-05-11
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    @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 $.2012-07-06
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    @RaymondManzoni You can answer http://math.stackexchange.com/questions/156372/is-it-possible-to-evaluate-int-01-xn-dx-by-contour-integration2012-07-06
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    @Yrogirg: I did just that (I had to change a bit the principle of the fixed angle $\frac {\pi}2$ to avoid the problems for odd $n$). I hope you'll like it!2012-07-06

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