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I would like to show that $\int_{-\infty}^\infty {\sin(ax)\exp(ibx)\over x}dx$ is equals to $\pi$ for $b\in (-a,a)$ and equals to $0$ otherwise.

So I thought I shall use the keyhole contour integration, with a positively orientated contour in the upper half plane and having an indentation at $z=0$. Since $z=0$ is a simple pole, I can apply the indentation theorem which gives me (I believe) an arc integral of the indentation as $0$ and since my contour does not enclose any singularities, Cauchy's theorem suggests that the contour integral is $0$. Now I need to find the integral of the larger arc (of radius $R$, say) of the contour and find its limit as $R\to \infty$. But I don't quite know how.

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Are you sure it's $b\in (-a,a)$ and not $b\in \{-a,a\}$? –  Dennis Gulko Mar 23 '12 at 9:57
    
Dear @DennisGulko, I am pretty sure that I want for $b\in (-a,a)$ the integral be $\pi$. :) –  Ken Mar 23 '12 at 10:11

1 Answer 1

up vote 2 down vote accepted

Consider the function $f(x)=\frac{\sin ax \,\,e^{ibx}}{x}=\frac{e^{i(b+a)x}-e^{i(b-a)x}}{2ix}$. Denote $t=b+a$ and $s=b-a$.
Define $g_t(z)=\frac{e^{itz}}{2iz}$. Now integrate this function on the upper half plane contour: $\gamma_1(t)=t, \,\,\gamma_2(t)=-t$ where $t\in[\varepsilon,R]$, and $\gamma_\varepsilon(t)=\varepsilon e^{i(\pi-t)},\,\,\gamma_R(t)=R e^{it}$ where $t\in[0,\pi]$.
What you need is $\lim_{R\to\infty,\, \varepsilon\to0}\int_{\gamma_1\oplus\gamma_2}(g_t(z)-g_s(z))dz$. Compute them separately - same way for both. It's easy to prove that $\int_{\gamma_R} g_t(z)dz\to0$ whenever $R\to\infty$ if $t> 0$. Also, $\int_{\gamma_\varepsilon}g_t(z)dz=-\pi i Res(g_t(z),0)$. Compute those for $t,s$ assuming each is positive, zero or negative, and you'll get the required result.

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Could you please explain the notation $\oplus$ in $\int_{\gamma_1\oplus\gamma_2}(g_t(z)-g_s(z))dz$? It seems to be the union (concatenation ?) of both line segments $\gamma_1$ and $\gamma_2$. –  Américo Tavares Mar 23 '12 at 10:36
    
Yes, that's what I meant, i.e. $\int_{\gamma_1\oplus\gamma_2}=\int_{\gamma_1}+\int_{\gamma_2}$. I didn't want to use $+$, to not get confused with the sum of the functions. –  Dennis Gulko Mar 23 '12 at 11:19
    
Thanks for the explanation! +1. –  Américo Tavares Mar 23 '12 at 12:02

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