Difficult integral involving exponential

zandria
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Homework Statement


I'm trying to verify the Fourier transform but am getting stuck on the integration. Here is the pair:
f(x) = e^{-ax^2}
\hat{f}(k) = \frac{1}{\sqrt{2a}}e^{-k^2/4a}
a>0

Homework Equations



I know that
\hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{ikx}dx

The Attempt at a Solution



So I have
\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2}e^{ikx}dx
\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2+ikx}dx

I tried using integration by parts and I'm not sure that's the right way to go. If it is I'm not sure how to go about it without getting a more complicated integral.
 
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You need to complete the square, which means:

-(ax^2+ikx)=-\left[(\alpha x+\beta)^2+\gamma \right].

Find \alpha,\beta and \gamma.

Edit1: it seems you have either listed \hat{f}(k) wrong or the book where you got \hat{f}(k) from is wrong, because the answer should be:

<br /> \hat{f}(k) = \sqrt{\frac{\pi}{a}}e^{-k^2/4a}<br />

edit2: While making no difference to the final answer in this case, shouldn't it be \hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, note the minus sign.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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