$F(k)=\frac{\sin{(k+1)}}{(k+1)}$.
How can I find the inverse Fourier Transform of above function.
I am trying to use the result:
$f(t)\exp(ikt)=F(k-k_0)$.
But I am stuck.
Please help me out. I am new to Fourier transform.
(inverse) Fourier transform of sinc function
fourier analysisfourier transform
Related Solutions
Substitute in the given expression. Then we are trying to calculate $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{2 + i\omega} e^{i\omega (t-1)} d\omega.$$
We can shift time to simplify the integrand: $$ f(t+1) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{2 + i\omega} e^{i\omega t} d\omega. $$
Now we flip through our favourite book of Fourier transforms and notice the following $$ f(t) = e^{-at} H(t); \quad \hat{f}(\omega) = \frac{1}{a + i\omega}.$$
Here $H(t)$ is the Heaviside step function. Excellent! Then clearly $$f(t+1) = e^{-2t}H(t).$$
And as a final step we undo the shift we introduced earlier, $$f(t) = e^{-2t-2}H(t-1).$$
The Gaussian $e^{-\alpha x^{2}}$ satisfies the differential equation $$ \frac{df}{dx} = -2\alpha x f,\;\;\; f(0)=1. $$ The Fourier transform turns differentiation into multiplication, and multiplication into differentiation. So you get back the same differential equation with different constants. For example, if $f(x)=e^{-\alpha x^{2}}$, then \begin{align} \frac{d}{dk}\hat{f}(k) & =\frac{d}{dk}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}(-ix)dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(-2\alpha xe^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\frac{d}{dx}e^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left(-\frac{d}{dx}\frac{i}{2\alpha}e^{-ikx}\right)dx \\ & = -\frac{k}{2\alpha}\hat{f}(k). \end{align} And, $$ \hat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx $$ Therefore, $$ \hat{f}(k) = \left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]e^{-k^{2}/4\alpha} $$ Determining the multiplier constant is normally done with a trick using polar coordinates: \begin{align} \left[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]^{2} & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-\alpha y^{2}}dxdy \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}e^{-\alpha r^{2}}rdrd\theta \\ & = \int_{0}^{\infty}e^{-\alpha r^{2}}rdr \\ & = -\frac{1}{2\alpha}\int_{0}^{\infty}\frac{d}{dr}e^{-\alpha r^{2}}dr \\ & = \frac{1}{2\alpha} \end{align} Finally, $$ \hat{f}(k) = \frac{1}{\sqrt{2\alpha}}e^{-k^{2}/4\alpha}. $$
Best Answer
Let me first get rid of the trivial shift, so I will consider what you call
$$ F(k-1) = \frac{\sin(k)}{k} = \mathrm{sinc}(k), $$
which is the sinc function.
The simplest way would be to recognize that the (inverse) Fourier transform of the box function is the sinc. This implies that the Fourier transform of $F(k-1)$ must be the box function. So the easiest thing would be to work backward and see what box exactly you need in order to get this particular expression for the sinc.
Let's instead work forward. We immediately face a problem as $F \notin L^1(\mathbb{R})$. This means that the integral
$$ \int_\mathbb{R} \ dk \ e^{ikx} \frac{\sin(k)}{k} $$
does not exist. What we should do is insert a cutoff $\Lambda$ after which the integral converges, and then evaluate the result as $\Lambda \to \infty$ :
$$ \lim_{\Lambda \to \infty} \int_{-\Lambda}^\Lambda dk \ e^{ikx} \frac{\sin(k)}{k} $$
However evaluating the integral with the cut-off is cumbersome so I'll use a different approach.
We are after
$$ f(x) = \int_\mathbb{R} \ dk \ e^{ikx} \frac{e^{ik}-e^{-ik}}{2ik} $$
where the integral must actually be intended in a peculiar way. Differentiating with respect to $x$ we obtain, formally,
\begin{align} f'(x) &= \int_\mathbb{R} \ dk \ e^{ikx} \frac{e^{ik}-e^{-ik}}{2} \\ &= \pi \left ( \delta(x+1)-\delta(x-1) \right ) \end{align}
where $\delta $ is the Dirac delta. Now we know that the primitive of the Dirac delta is the Heaviside function $\theta(x)$ so, integrating with respect to $x$ we get
$$ f(x) = \pi \left ( \theta(x+1)-\theta(x-1) \right ). $$
This is precisely the box function between -1 and 1 and height $\pi$.
Introducing back the shift we get
$$ \hat{F}(x) = \pi \left ( \theta(x+1)-\theta(x-1) \right ) e^{-ix} $$