Question

Find the Fourier Cosine transform of e^-x​

Answer 1

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To reiterate, call I=∫∞0e−x2cos(sx)dx. Now, dIds=∫∞0−xe−x2sin(sx)dx=12∫∞0sin(sx)(−2xe−x2)dx. Use integration by parts to get:12[sin(sx)e−x2∣∣∞0−s∫∞0cos(sx)e−x2dx]=−s2∫∞0e−x2cos(sx)dx=−s2I. So we have a differential equation: dII=−∫s2ds+log(c). Hence, log(I)=−s24+log(c)=log(ce−s24). Solving for I, we get that I=ce−s24=∫∞0e−x2cos(sx)dx. I'll leave you to solve for c, just plug in s=0 and then you get c=...

Answer 2

Answer:

The Fourier cosine transform of the function $e^{-x}$ is $\dfrac{1}{1+s^2}$.

Step-by-step explanation:

The question is to find the Fourier cosine transform of the function $e^{-x}$. The formula to find Fourier cosine transform of f(x) is given by,

$f_{c}\left( s\right) =\int ^{\infty }_{0}f\left( x\right) \cos \left( sx\right) dx$

Here is the given $f\left( x\right) =e^{-x}$. Then by definition of Fourier cosine transform of f(x),

$f_c\left(s\right)=\int_0^{\infty}e^{-x}\cos\left(sx\right)dx$

Now we know that the standard form of $\int_{ }^{ }e^{ax}\cos\left(bx\right)dx$ is given by,$\int_{ }^{ }e^{ax}\cos\left(bx\right)dx=\dfrac{e^{ax}}{a^2+b^2}\left(a\cos bx+b\sin bx\right)$

Here we have $a=-1$ and $b=s$. Then,

$f_{c}\left( s\right) =\dfrac{e^{-x}}{1+s^{2}}\left[ -\cos sx+s\sin sx\right] _{0}^{\infty }$

By applying lower limit and upper limit we get,

$\begin{aligned}f_c\left(s\right)=\left[\frac{e^{-\infty}}{1+s^2}\left(-\cos\infty+s\sin_{\infty}\right)\right]-\end{aligned}\left[\dfrac{e^0}{1+s^2}\left(-\cos 0+s\sin0\right)\right]$

We know that , $e^{0} =1$, $cos (0)=1$, and $sin(0)=0$. By substituting these values we get,

$f_c\left(s\right)=0-\dfrac{1}{1+s^2}\left( -1+0\right)$

And the final answer of the Fourier cosine transform of the function $e^{-x}$ is,

$f_c\left(s\right)=\dfrac{1}{1+s^2}$