Question

Find the Fourier sine transform of - 2e^-3x + 3e^-2x​

Answer 1

Answer:

he Fourier sine transform is the imaginary part of the full complex Fourier transform,

F_x^((s))[f(x)](k) = I[F_x[f(x)](k)]  

(1)

= int_(-infty)^inftysin(2pikx)f(x)dx.  

(2)

The Fourier sine transform F_s(k) of a function f(x) is implemented as FourierSinTransform[f, x, k], and different choices of a and b can be used by passing the optional FourierParameters -> {a, b} option. In this work, a=0 and b=-2pi.

Step-by-step explanation:

Answer 2

Answer:

The Fourier sine transform of the given function is

$\sqrt{\dfrac{2}{\pi } }\bigg[{{\dfrac{3s}{4 +s^{2} }-\dfrac{2s}{9 +s^{2} } } }} }\bigg]$

Step-by-step explanation:

Given a function  

$f(x)=-2e^{-3x} + 3e^{-2x}$

The Fourier sine transform of a function is given by,

$F_s[f(x)]=\sqrt{\dfrac{2}{\pi } } {\displaystyle \int\limits^{\infty}_0 {f(x)sinsx } \, dx$

Therefore, for the given function, the Fourier sine transform is

$F_s[f(x)]=\sqrt{\dfrac{2}{\pi } } {\displaystyle\int\limits^{\infty}_0 {\big(-2e^{-3x} + 3e^{-2x}\big)sinsx } \, dx$

             $=\sqrt{\dfrac{2}{\pi } }\bigg[ {\displaystyle\int\limits^{\infty}_0 {-2e^{-3x} sinsx } \, dx~+ {\displaystyle\int\limits^{\infty}_0 { 3e^{-2x}sinsx } \, dx\bigg]$

             $=-2\sqrt{\dfrac{2}{\pi } } {\displaystyle\int\limits^{\infty}_0 {e^{-3x} sinsx } \, dx~+3\sqrt{\dfrac{2}{\pi } } {\displaystyle\int\limits^{\infty}_0 { e^{-2x}sinsx } \, dx$

Using the integral,

${\displaystyle\int {e^{ax} sinbx } \, dx~={\dfrac{e^{ax}}{a^{2} +b^{2} } (a sinbx -bcosbx)} }$

$F_s[f(x)]= -2\sqrt{\dfrac{2}{\pi } }\bigg[{\dfrac{e^{-3x}}{(-3)^{2} +s^{2} } (-3 sinsx -scossx)} ~}\bigg]\limits^{\infty}_0 \\~~~~~~~~~~~~~~~~~~~~~~~+3\sqrt{\dfrac{2}{\pi } }\bigg[{\dfrac{e^{-2x}}{(-2)^{2} +s^{2} } (-2 sinsx -scossx)} ~}\bigg]\limits^{\infty}_0$

Since, $e^{\infty}}=0$ and $e^{0}=1$, we get

$F_s[f(x)]=-2\sqrt{\dfrac{2}{\pi } }\bigg[{[0]-\dfrac{1}{9 +s^{2} } (-s)} ~}\bigg] +3\sqrt{\dfrac{2}{\pi } }\bigg[{[0]-\dfrac{1}{4 +s^{2} } (-s)} ~}\bigg]$

            $=-2\sqrt{\dfrac{2}{\pi } }\bigg[{\dfrac{s}{9 +s^{2} } } }\bigg] +3\sqrt{\dfrac{2}{\pi } }\bigg[{\dfrac{s}{4 +s^{2} }} }\bigg]$

            $=\sqrt{\dfrac{2}{\pi } }\bigg[{\dfrac{-2s}{9 +s^{2} } } }+{\dfrac{3s}{4 +s^{2} }} }\bigg]$

           $=\sqrt{\dfrac{2}{\pi } }\bigg[{{\dfrac{3s}{4 +s^{2} }-\dfrac{2s}{9 +s^{2} } } }} }\bigg]$

Therefore, the Fourier sine transform of the given function is $\sqrt{\dfrac{2}{\pi } }\bigg[{{\dfrac{3s}{4 +s^{2} }-\dfrac{2s}{9 +s^{2} } } }} }\bigg]$

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