Compute the following integral given the Fourier transform of $f$ I'm using given the following definition of the Fourier transform: $$\hat{f}(k)=F[f]=\int_{\mathbb{R}} f(x)e^{-ikx} \, dx $$
I'm then asked to find the transform of the function defined by $$\begin{cases} e^{-x} & x>0 \\ 0 & x<0\end{cases}$$
I have done this and got $$F[f]=\frac{1}{1+ik}$$
I'm then asked to use the inversion formula $$F^{-1}[\hat{f}]=\frac{1}{2\pi}\int_{\mathbb{R}} \hat{f}(k)e^{ikx} \, dk $$ to compute $$\int_{0}^{\infty}\frac{\cos kx+k\sin kx}{1+k^2} \, dk$$
How do I do this part?
 A: We have the Fourier Transform pair
$$f(x)=e^{-x}u(x)\leftrightarrow F(k)=\frac{1}{1+ik}$$
Denote the unit step function by $u(x)$.  Then, we see that 
$$\begin{align}
e^{-x}u(x)&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{e^{-ikx}}{1+ik}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{e^{-ikx}}{1+ik}\frac{1-ik}{1-ik}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{e^{-ikx}(1-ik)}{1+k^2}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{(\cos(kx)-i\sin(kx))(1-ik)}{1+k^2}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\cos(kx)+k\sin(kx)}{1+k^2}\,dk-i\frac{1}{2\pi}\underbrace{\int_{-\infty}^\infty\frac{k\cos(kx)+\sin(kx)}{1+k^2}}_{=0\,\,\text{since we have an odd integrand}}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\cos(kx)+k\sin(kx)}{1+k^2}\,dk
\end{align}$$
Therefore, we find that 

$$\int_{-\infty}^\infty\frac{\cos(kx)+k\sin(kx)}{1+k^2}\,dk=2\pi e^{-x}u(x)$$

A: Couple of hints:
$e^{ik} = \cos k + i\sin k$
and $F[f]$ can be re-written
$\displaystyle \frac {1-ik}{1+k^2}$
