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I have the integral

${\operatorname{Im}} \left (\int^\infty_0 e^{ix} x^{s-1} \, \mathrm{d} x \right)$

and I wish to redefine $x \to iy$ but I am unsure of how to justify this using contour integration arguments. I would be very grateful for any help. Thanks.

Max Clifford
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1 Answers1

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An idea:

$$\int\limits_0^\infty e^{ix}x^{s-1}dx=\int\limits_0^\infty \left(x^{s-1}cos x+ix^{s-1}\sin x\right)dx\implies$$

$$\implies\operatorname{Im}\left(\int\limits_0^\infty e^{ix}x^{s-1}dx\right)=\int\limits_0^\infty x^{s-1}\sin x\,dx$$

The last integral converges for $\,s<1\,$

DonAntonio
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  • DonAntonio, thank you for your reply. It was very kind of you but unfortunately I already know that - I am following a proof in a textbook and the author skips a step or two along the way. I need a justification for the redefinition above. I am not actually trying to evaluate the integral or look at convergence properties. Thanks – Max Clifford Apr 09 '13 at 03:19
  • Anytime, @MaxClifford . So you want to substitute $,x\to iy,$ , but that will give you $,i\infty,$ as upper limit in the integral, am I right? And still I don't understand then what you mean by "contour integration arguments": what contour? You have a real integral of a complex function which, I suppose, you can try to evaluate by definition as a line integral. – DonAntonio Apr 09 '13 at 03:41