What is the Fourier cosine transform of $e^{-ax}$
I got
$$
\int_{0}^{\infty}\cos(kx)e^{-ax}dx = \frac{e^{-ax}(k\sin(kx) -\cos(kx))} {a^{2}+k^{2}}\Bigr|_{0}^{\infty}
$$
But how do you continue from here?
[Math] what is the Fourier cosine transform of $e^{-ax}$
calculusfourier analysisfourier transformintegrationreal-analysis
Related Solutions
It usually helps to follow Fourier's prescription rather than jump to the solution. For Laplace's equation, the separated solutions $X(x)Y(y)$ satisfy $$ X''Y+XY'' = 0, \\ \frac{X''}{X} = -\frac{Y''}{Y} = \lambda, $$ where $\lambda$ is a separation parameter. So the Fourier ODEs are $$ \begin{align} X''=\lambda X, &\;\;\;\;\; Y''=-\lambda Y \\ 0 < x < a, & \;\;\;\;\;0 < y < \infty \\ X(0) = 0, & \;\;\;\;\; Y'(0) = 0. \end{align} $$ Fourier always assumed the separated solutions were bounded on any unbounded domain. The conditions for $X$ do not determine parameters $\lambda$; however the boundedness of $Y$ in conjunction with $Y'(0)=0$ forces $\lambda \ge 0$. (If $\lambda < 0$ then $Y(y)=\cosh(\sqrt{\lambda}y)$ is unbounded.) Therefore, $\lambda \ge 0$ is assumed, and the separated solutions for $\lambda=\mu^{2}$ satisfying the above conditions are $$ X_{\mu}(x)Y_{\mu}(y) = C(\mu)\sinh(\mu x)\cos(\mu y). $$ A sum of these must be an integral sum: $$ u(x,y) = \int_{0}^{\infty}C(\mu)\sinh(\mu x)\cos(\mu y)dy. $$ The coefficient function $C(\mu)$ is the only missing component, and must be chosen so that $$ g(y) = u(a,y) = \int_{0}^{\infty}C(\mu)\sinh(\mu a)\cos(\mu y)dy. $$ Writing $g$ as a Fourier cosine transform gives $$ g(y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}\cos(\mu u)g(u)du\right)\cos(\mu y)d\mu, $$ which leads to $$ C(\mu)\sinh(\mu a) = \frac{2}{\pi}\int_{0}^{\infty}\cos(\mu u)g(u)du \\ C(\mu) = \frac{2}{\pi\sinh(\mu a)}\int_{0}^{\infty}\cos(\mu u)g(u)du. $$ The final Fourier solution is $$ u(x,y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\frac{1}{\sinh(\mu a)} \int_{0}^{\infty}\cos(\mu u)g(u)du\right)\sinh(\mu x) \cos(\mu y)d\mu. $$ Sanity Check: Note that the expression in parentheses is a function of $\mu$ only--it is $C(\mu)$. Visual inspection of the final, proposed solution shows that $u(a,y)$ is the inverse Fourier cosine transform of the Fourier cosine transform of $g$. So $u(a,y)=g(y)$ under suitable smoothness conditions on $g$. Cleary $u(0,y)=0$. Under suitable assumptions to ensure the convergence of the differentiated integral expression, $u_{y}(x,0)=0$ follows. All of the required conditions check, and this does give a solution of Laplace's equation given sufficient convergence of the different derived series.
The last line is wrong, it's supposed to be: $$\frac{\sin(2\pi u)}{2\pi}\Big(\frac{1}{u}-\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$$ I'll work out the first term, the rest are similar: $$F(\omega)=\mathcal{F}_u\bigg[\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)\bigg]=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du$$
Keep in mind that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Hence: $$F(\omega)=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du=\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega-2\pi) u}du-\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega+2\pi) u}du$$
Using two facts $$\mathcal{F}_t[e^{i\omega_0t}\cdot f(t)]=F(\omega-\omega_0)$$ where $G(\omega)=\mathcal{F}_t[f(t)]$ and $$\mathcal{F}_t\bigg[\frac 1t\bigg]=-i\pi\text{sgn} (\omega)$$
You get $$F(\omega)=-\frac 1{4} \text{sgn}(\omega-2\pi)+\frac 1{4} \text{sgn}(2 \pi+\omega)$$
For the other parts, using the following property (very easy to prove): $$\mathcal{F}_t(f(t-t_0))=e^{-it_0\omega}F(\omega)$$ $$F_1(\omega)=\frac 12 e^{-i\omega/2}F(\omega)$$ $$F_2(\omega)=\frac 12 e^{i\omega/2}F(\omega)$$
The transform $G_u(\omega)$ of the function $f(u)$ is: \begin{align*} G_u(\omega)&=F(\omega)+F_1(\omega)+F_2(\omega)\\ &=(1+\cos(w/2)F(\omega)\\ &=2\cos^2(\omega/4)F(\omega) \end{align*}
So that $$\mathcal{F}_t(g(t))=G(\omega)=\frac T2 \cos^2(\frac{T\omega} 4) \left(\text{sgn}(2 \pi - T\omega) + \text{sgn}(2 \pi + T\omega)\right)$$
Best Answer
The expression you obtained is incorrect. It is missing a factor of $a$ in front of the cosine term.
Assume $a > 0$. (Otherwise the integral does not converge.) \begin{align} \int_{0}^{\infty}\cos(k x)e^{-ax}\,dx &= Re \int_{0}^{\infty}e^{i k x}e^{-ax}\,dx\\ &= Re \int_{0}^{\infty}e^{(i k - a)x}\,dx\\ &= Re\; \frac{e^{(i k - a)x}}{i k - a}\; \Bigr|_{0}^{\infty} \end{align} Since $a>0$, $\lim_{x\rightarrow\infty}e^{(i k - a)x}$ is zero, and we are left with: $$ \int_{0}^{\infty}\cos(k x)e^{-ax}\,dx \;=\; Re\; \frac{-1}{i k - a} \;=\; \frac{a}{a^2 + k^2} $$
As weird as it may sound for a question which is evidently purely-mathematical, one way to catch an error such as the missing $a$ above is by dimensional analysis. Imagine that the units of $x$ are length. Then for consistency, $a$ and $k$ must have units of inverse length. That means you can't add or subtract two terms like $k \sin(k x)$ and $\cos(kx)$, since they have different units. (One has units of inverse length, the other is dimensionless.) If you arrive at such an expression, then there must have been a mistake somewhere.