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I am considering the integral of $z^{s-1}e^{iz}$ (where $0<s<1$) around a contour that is a triangle with vertices at 0, $R$, and $iR$, except for a circular cutout around the origin (i.e. after going down the positive imaginary axis from $iR$ to $i\delta$, go along an arc counterclockwise to $\delta$ on the positive real axis). I am trying to show that the integral along the slanted line (from $R$ to $iR$) approaches 0 as $R\rightarrow\infty$.

I used the usual approximation of the modulus of the integral by the maximum modulus of the integrand on the path times the length of the path but it wasn't giving me an upper bound that was approaching 0. I am wondering what other method I should be using or whether the integral does approach 0?

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What is the range of $s$? – 23rd Nov 18 '12 at 8:58
@richard: I'm sorry, I forgot to include that. $0<s<1$. – user44532 Nov 18 '12 at 16:22
up vote 0 down vote accepted

You have the constraint $\,x+y=R\,$ (with $z:=x+iy$) so that you want : $$I(R)=(1-i)\int_0^R (x+i(R-x))^{s-1}\;e^{i(x+i(R-x))}\,dx$$ An upper bound for $|I(R)|$ is : $$|I(R)|\le\sqrt{2}\int_0^R \left(x^2+(R-x)^2\right)^{\frac {s-1}2}\;e^{x-R}\,dx$$

Let's suppose that $\,s< 1$ (for $s>1$ the initial integral is not bounded as $R\to \infty$).
We observe that $\ x^2+(R-x)^2=2\bigl(x-\frac R2\bigr)^2+\frac {R^2}2\ $ so that $\ \frac{R^2}2\le x^2+(R-x)^2\le R^2\ $ and :

$$|I(R)|\le\sqrt{2}\left(\frac{R^2}2\right)^{\frac {s-1}2}\int_0^R e^{x-R}\,dx$$ getting : $$|I(R)|\le K(s)R^{s-1}\left(1-e^{-R}\right)$$ and $$\boxed{\displaystyle\lim_{R\to+\infty} |I(R)|=0}\quad \text{as wished}$$

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Is there a reason why you consider a slanted line (from $R$ to $\imath R$)? If you would replace the line by a quadrant, you could apply Jordan's lemma... in this case you only have to show that $\sup_{|z|=R} |z^{s-1}| \to 0$ as $R \to \infty$.

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