Question

how to find the fourier sine and cosine tranforms of the function ...
1.x^(n-1)
2.1/(1+x^2)

Answer 1

I don't think the first function has either transform. The integrals would only converge for \(n=1\). Unless there's something I'm missing... in which case you can attempt to establish a recurrence relation after several rounds of integration by parts.

For the second, consider using the Laplace transform. Notice that \(\dfrac{1}{1+x^2}\sin(2\pi nx)\) is an odd function for all \(n\), so the sine transform should disappear.
\[\mathcal{F}_s\left(\frac{1}{1+x^2}\right)=I_n=\int_{-\infty}^\infty \frac{\sin(2\pi nx)}{1+x^2}\,dx\]
Taking the Laplace transform, we have
\[\begin{align*}\mathcal{L}\{I_n\}&=\int_{-\infty}^\infty \int_0^\infty\frac{\sin(2\pi nx)}{1+x^2}e^{-sn}\,dn\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\int_0^\infty \sin(2\pi nx)e^{-sn}\,dn\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\mathcal{L}\{\sin(2\pi nx)\}\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\cdot\frac{2\pi x}{s^2+(2\pi)^2x^2}\,dx\\\\
&=\frac{2\pi}{s^2-(2\pi)^2}\int_{-\infty}^\infty \left(\frac{x}{x^2+1}-\frac{(2\pi)^2x}{s^2+(2\pi)^2x^2}\right)\,dx\\\\
&=0\end{align*}\]
Meanwhile, the cosine transform does not disappear:
\[\mathcal{F}_c\left(\frac{1}{1+x^2}\right)=J_n=\int_{-\infty}^\infty \frac{\cos(2\pi nx)}{1+x^2}\,dx\]
Taking the Laplace transform,
\[\begin{align*}\mathcal{L}\{J_n\}&=\int_{-\infty}^\infty \int_0^\infty\frac{\cos(2\pi nx)}{1+x^2}e^{-sn}\,dn\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\int_0^\infty \cos(2\pi nx)e^{-sn}\,dn\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\mathcal{L}\{\cos(2\pi nx)\}\,dx\\\\
&=\int_{-\infty}^\infty \frac{1}{1+x^2}\cdot\frac{s}{s^2+(2\pi)^2x^2}\,dx\\\\
&=\frac{s}{s^2-(2\pi)^2}\int_{-\infty}^\infty \left(\frac{1}{1+x^2}-\frac{(2\pi)^2}{s^2+(2\pi)^2x^2}\right)\,dx\\\\
&=\frac{s}{s^2-(2\pi)^2}\left(\pi-\frac{2\pi^2}{s}\right)\\\\
J_n&=\mathcal{L}^{-1}\left\{\frac{\pi s}{s^2-(2\pi)^2}-\frac{2\pi^2}{s^2-(2\pi)^2}\right\}\\\\
&=\pi\cosh(2n\pi)-\pi\sinh(2n\pi)
\end{align*}\]

Answer 2

i think the first function will have fourier transform
proceed with \[F_{sine}(x^{n-1})=\sqrt{\frac{ 2 }{ \pi }}\int\limits_{0}^{\infty}\sin(sx).x^{n-1}dx\\F_{\cos}(x^{n-1})=\sqrt{\frac{ 2 }{ \pi }}\int\limits_{0}^{\infty}\cos(sx).x^{n-1}dx\]

Answer 3

now add them up
\[F_{\cos}(x^{n-1})+F_{sine}(x^{n-1})=\frac{ 2 }{ \sqrt{\pi} }\int\limits_{0}^{\infty}(cossx+i.sinsx)x^{n-1}dx\\=\sqrt{\frac{ 2 }{ \pi}}\int\limits_{0}^{\infty}e^{isx}.x^{n-1}dx\\now~put,~isx=-t\\then ~,x=\frac{ -t }{ is }\\dx=\frac{ -dt }{ is }\]

Answer 4

so the integral is now\[\sqrt{\frac{ 2 }{ \pi }}\int\limits_{0}^{\infty}e^{-t}(\frac{ -t }{ is })^{n-1}(\frac{ -dt }{ is})\\=\sqrt{\frac{2}{\pi}}\frac{ 1 }{ (is)^n }(-1)^{n} \int\limits_{0}^{\infty}e^{-t}t^{n-1}\\the~Integral~is ~called ~a~ \Gamma~function\\=\sqrt{\frac{ 2 }{ \pi }}\frac{ (i)^{2n} }{ (i)^{n}.s^{n} }*\Gamma n=\sqrt{\frac{ 2 }{ \pi }}\frac{ (i)^n }{ s^{n} }\Gamma n\\=\sqrt{\frac{ 2 }{ \pi }}\frac{ \cos \frac{ n \pi }{ 2 } +i.\sin \frac{ n .\pi}{ 2 }}{ s^{n}}.\Gamma n\]
now equate real and imaginary parts
u'll get-

Answer 5

\[F_{\cos}(x^{n-1})=\sqrt{\frac{2}{\pi}} \frac{\Gamma n}{s^{n}}\cos\frac{n \pi}{2}\\and \\F_{sine}(x^{n-1})=\sqrt{\frac{2}{\pi}} \frac{\Gamma n}{s^{n}}\sin\frac{n \pi}{2}\]
i'm not sure about it @SithsAndGiggles please take a look :)

Answer 6

and for the second problem follow the same Strategy

Answer 7

but theres a very easy method to solve the second one first find out fourier sine and cosine transforms of e^(-x)\[F_{\cos}(e^{-x})=\sqrt{\frac{2}{\pi}}\frac{1}{s^2+1}\\now just~ inverse~fourier ~\it\\replace ,s=x~In~Right~Hand~Side ~and ~x=s~In~Left~ hand~side \\e^{-s}=\sqrt{\frac{2}{\pi}}*F_{\cos}(\frac{1}{1+x^2})\\\\F_{\cos}(\frac{1}{1+x^2})=\sqrt{\frac{\pi}{2}}e^{-s}\]
and same procedure for sine ,this is a shortcut method if u find difficulty then use the conventional integration process

Answer 8

I'm thoroughly confused... Several sources I've checked with use the definition I'm using in my work:
\[\large\mathcal{F}\{f(x)\}=\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\xi x}\,dx=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\,dx\]
where I use \(n\) in place of \(\xi\) for the second function.

I checked the definition used in Mathematica's computation:
\[\large\mathcal{F}\{f(x)\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{i\xi x}\,dx\]
which is used in this file (page 4): https://web2.ph.utexas.edu/~gleeson/ElectricityMagnetismAppendixB.pdf

On that same page the author lists the definitions for the since and cosine transforms that you use @sidsiddhartha (equations B.15 and B.16):
\[\large\mathcal{F}_{\cos}\{f(x)\}=\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos\xi x \,dx\\
\large\mathcal{F}_{\sin}\{f(x)\}=\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin\xi x \,dx\]
It's been some time since I learned about the transform, so I'm probably forgetting some of the essential properties. I will have to do some reviewing on the topic...

Answer 9

Also, these were the definitions I was using for the sine and cosine transforms: http://en.wikipedia.org/wiki/Sine_and_cosine_transforms