indefinite integral

**hurz** · June 22nd 2011, 04:04 PM

How to evaluate
$\int_{-\infty}^{\infty} \exp(-x^2) \exp(-i\alpha x) dx = \int_{-\infty}^{\infty} \exp(-x^2-i\alpha x) dx$
, the Fourier transform of $\exp(-x^2)$
Thanks

**Ackbeet** · June 22nd 2011, 04:27 PM

Try completing the square inside the exponent.

**chisigma** · June 22nd 2011, 11:57 PM

Originally Posted by hurz

How to evaluate
$\int_{-\infty}^{\infty} \exp(-x^2) \exp(-i\alpha x) dx = \int_{-\infty}^{\infty} \exp(-x^2-i\alpha x) dx$
, the Fourier transform of $\exp(-x^2)$
Thanks

First we define...

$f(x)= e^{-x^{2}}$ (1)

$F(\alpha)= \mathcal{F} \{f(x)\}= \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx$ (2)

... and then, remembering that is...

$\mathcal{F} \{f^{'} (x) \}= i\ \alpha\ \mathcal{F} \{f(x)\}$ (3)

... from (1), (2) and (3) we derive...

$F^{'} (\alpha) = - \frac{\alpha}{2}\ F(\alpha)$ (4)

The (4) is an ordinary first order DE and its general solution is found separating the variables...

$\frac{d F}{F} = - \frac{\alpha}{2}\ d \alpha \implies \ln F = - \frac{\alpha^{2}}{4} + \ln c \implies F = c\ e^{- \frac{\alpha^{2}}{4}}$ (5)

Taking into account the known result $F(0)= \sqrt {\pi}$ we obtain the final result...

$\int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt {\pi}\ e^{- \frac{\alpha^{2}}{4}}$ (6)

Kind regards

$\chi$ $\sigma$

**hurz** · June 23rd 2011, 02:27 PM

Originally Posted by chisigma

First we define...

$f(x)= e^{-x^{2}}$ (1)

$F(\alpha)= \mathcal{F} \{f(x)\}= \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx$ (2)

... and then, remembering that is...

$\mathcal{F} \{f^{'} (x) \}= i\ \alpha\ \mathcal{F} \{f(x)\}$ (3)

... from (1), (2) and (3) we derive...

$F^{'} (\alpha) = -2\ \alpha\ F(\alpha)$ (4)

The (4) is an ordinary first order DE and its general solution is found separating the variables...

$\frac{d F}{F} = -2\ \alpha\ d \alpha \implies \ln F = - \alpha^{2} + \ln c \implies F = c\ e^{- \alpha^{2}}$ (5)

Taking into account the known result $F(0)= \sqrt {\pi}$ we obtain the final result...

$\int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt {\pi}\ e^{- \alpha^{2}}$ (6)

Kind regards

$\chi$ $\sigma$

I'm sorry, can you ive details about how to derive (4) ?
What does $F^{'} (\alpha)$ mean? The derivate of the fourier transform?
Thanks

**chisigma** · June 23rd 2011, 10:48 PM

Originally Posted by hurz

I'm sorry, can you ive details about how to derive (4) ?

What does $F^{'} (\alpha)$ mean? The derivate of the fourier transform?

Thanks

All right!... that's useful because there is an error in my previous post!... it is...

$\mathcal{F} \{f^{'}(x)\} = - 2\ \int_{- \infty}^{+ \infty} x\ e^{-x^{2}}\ e^{-i \alpha x}\ dx =$

$= - 2\ i \frac{d}{d \alpha}\ \int_{- \infty}^{+ \infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = - 2\ i\ F^{'} (\alpha)$ (1)

... so that from (1)...

$F^{'} (\alpha) = \frac{1}{2 i}\ \mathcal{F} \{f^{'}(x)\} = - \frac{\alpha}{2}\ F(\alpha)$ (2)

... and the (2) with the 'initial condition' $F(0)= \sqrt{\pi}$ gives the result...

$\int_{- \infty}^{+ \infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt{\pi}\ e^{-\frac{\alpha^{2}}{4}}$ (3)

Sorry for my previous error!...

Kind regards

$\chi$ $\sigma$

Thread: indefinite integral

LinkBack

Thread Tools

Display