what is the Fourier cosine transform of $e^{-ax}$ 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?
The expression you obtained is incorrect. It is missing a factor of $a$ in front of the cosine term.

Assume $a &gt; 0$. (Otherwise the integral does not converge.)
\begin{align}
\int_{0}^{\infty}\cos(k x)e^{-ax}\,dx &amp;= Re \int_{0}^{\infty}e^{i k x}e^{-ax}\,dx\\
&amp;= Re \int_{0}^{\infty}e^{(i k - a)x}\,dx\\
&amp;= Re\; \frac{e^{(i k - a)x}}{i k - a}\; \Bigr|_{0}^{\infty}
\end{align}
Since $a&gt;0$, $\lim_{x\rightarrow\infty}e^{(i k - a)x}$ 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.
Observe that if $a&gt;0$ then
$$
\lim_{x\to+\infty}e^{-ax}(k\sin(kx) -\cos(kx))=0.
$$