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I'm trying to compute the Fourier transform of the function $u_y(x)=e^{-y|x|}$ and i've come up to calculating $$I=\int_0^{\infty}e^{-(i\xi+y)x}dx$$ for example. Can i say this integral is $$\frac{e^{-(i\xi+y)x}}{i\xi+y}$$ 'evaluated' from zero to $\infty$ ?

I thought this: considering the curve $\gamma:[0,\infty[\longrightarrow \mathbb{C}$ given by $\gamma(x)=(i\xi+y)x$ i have $I=\frac{1}{i\xi+y}\int_{\gamma}e^{-z}dz$ from the definition of integral along a parametrized path. So now i should say that $\int_{\gamma}e^{-z}dz=\int_0^{\infty}e^{-x}dz=1$. I should integrate over a closed path composed by $\gamma_1:[0,R] \longrightarrow \mathbb{C}$, $\gamma_1(x)=x$, $\gamma_2$ a part of a circle of radius R starting from $(R,0)$ up to the intersection with the curve $\gamma$. If i prove the integral over $\gamma_2$ tends to zero as R tends to $\infty$ i will have proved my assertion, but i can't do this. This is what i got: $\gamma_2(x)=Re^{ix}$ with domain $[0,T]$ for some T. Inserting into the integral (call it J) i come up with the estimate $$J \leq R\int_0^T|e^{Re^{ix}}|dx =R\int_0^Te^{R\cos{x}}$$ but it doesn't seem to approach zero as R goes to $\infty$!

(Anyway, if this is correct i get $\frac{2y}{y^2+\xi^2}$ as the fourier transform)

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Why are you introducing a circular arc out of nowhere? You should have just evaluated at $0$ and $\infty$ (though your antiderivative has a sign error); the fundamental theorem of calculus is still valid. –  anon Jun 23 '12 at 22:49

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The function to integrate is $u:z\mapsto\mathrm e^{-z}$ and the path is $\gamma_2$ with equation $\gamma_2(x)=R\mathrm e^{\mathrm ix}$ for $x$ in $[0,T]$. Furthermore, the original parameter $y$ is positive hence $0\leqslant T\lt\frac\pi2$.

Note that $u(\gamma_2(x))=\mathrm e^{-R\mathrm \exp(\mathrm ix)}$ hence $|u(\gamma_2(x))|=\mathrm e^{-R\cos(x)}$ (you might have forgotten a minus sign here). Using this in the evaluation of $J$ and the fact that $\cos(x)\geqslant \cos(T)$ uniformly over $x$ in $[0,T]$, one gets $$ |J|\leqslant R\int_0^T\mathrm e^{-R\cos(x)}\,\mathrm dx\leqslant RT\mathrm e^{-R\cos(T)}. $$ The limit of the RHS is $0$ when $R\to+\infty$ because $\cos(T)\gt0$.

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