Question

Fourier sine transform of 1/x​

Answer 1

I hope it will help you....

Answer 2

Answer:

$I=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{s}{a} ) ]+c$

Step-by-step explanation:

For initial condition, putting s=0, then c=0

Therefore, from (2), we have

$I=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ] = > F{f(x)}=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ]$

$F(s)=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ] = > F{{f(x)}}=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ]$

$F(s)=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ]$

$f_{x} (\frac{e^{-ax} }{x} )=\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ]$

$\sqrt{\frac{2}{π} }\int\limits^0_x {\frac{e^{-ax} }{x} } \,sinx(sx) dx =\sqrt{\frac{2}{π} } [tan^{-1}(\frac{x}{a} ) ]$

#SPJ3