Question

the fourier cosine transform of f(x) = 2e^-5x+5e^-2x​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given: $f(x) = 2e^{-5x} + 5e^{-2x}$

To Find: Fourier Cosine Transform of $f(x) = 2e^{-5x} + 5e^{-2x}$

Solution:

• Finding Fourier Cosine Transform (FCT) of the given function

Fourier Cosine Transform of the function $f(x)$ is given by,

$f_{c}(s) = \sqrt{\frac{2}{\pi}} \int\limits^{\infty}_0 {f(x) cos(sx)} \, dx$

Therefore, for the function $f(x) = 2e^{-5x} + 5e^{-2x}$, Fourier Cosine Transform is given by;

$\Rightarrow f_{c}(s) = \sqrt{\frac{2}{\pi}} \int\limits^{\infty}_0 {(2e^{-5x} + 5e^{-2x}) cos(sx)} \, dx$

$\Rightarrow f_{c}(s) = \sqrt{\frac{2}{\pi}} \int\limits^{\infty}_0 {(2e^{-5x}) cos(sx)} \, dx + \sqrt{\frac{2}{\pi}} \int\limits^{\infty}_0 {(5e^{-2x}) cos(sx)} \, dx$

The terms $(2e^{-5x}) cos(sx)$ and $(5e^{-2x}) cos(sx)$ can either be integrated by using the 'by-parts' integration method or the following formula can be used to get the integral value:

$\int\limits^{\infty}_0 {e^{ax} cos(bx) } \, dx = \frac{e^{ax}}{a^{2} + b^{2} } (a \ cos(bx) + b \ sin(bx))$

Therefore,

$\Rightarrow f_{c}(s) = \frac{2\sqrt{2}}{\pi} [ \frac{e^{-5x} (-5 \ cos(sx) + s \ sin(sx))}{(-5)^{2} + s^{2} } ]_{_{0}}^{^{\infty}} + \frac{5\sqrt{2}}{\pi} [ \frac{e^{-2x} (-2 \ cos(sx) + s \ sin(sx))}{(-2)^{2} + s^{2} } ]_{_{0}}^{^{\infty}}$

$\Rightarrow f_{c}(s) = \frac{2\sqrt{2}}{\pi} [ 0 - \frac{(-5)}{25 + s^{2} } ]+ \frac{5\sqrt{2}}{\pi}[ 0 - \frac{(-2)}{4 + s^{2} } ]$

$\Rightarrow f_{c}(s) = \frac{10\sqrt{2}}{\pi} [\frac{1}{25 + s^{2} } + \frac{1}{4 + s^{2} } ]$

Hence, the Fourier Cosine Transform of $2e^{-5x} + 5e^{-2x}$ is

$\Rightarrow f_{c}(s) = \frac{10\sqrt{2}}{\pi} [\frac{1}{25 + s^{2} } + \frac{1}{4 + s^{2} } ]$