How does one integrate $w(x)=\int_{-\infty}^{\infty}\frac{1}{1+2\pi i\xi}e^{2\pi i\xi x}d\xi$ ?
This should be $w(x)=e^{-x}u(x)$ where $u(x)$ is the Heaviside unit step function.
In other words, for $x<0$, $w(x)=0$, and for $x\geq 0$, $w(x)=e^{-x}$.
I want to know how to do the integral. Thank you for teaching me!
I thought it might be instructive to present a solution that relies on classical real analysis only. We will augment the Fourier transform expression so that we may rigorously differentiate under the integral. Proceeding, we have
$$\begin{align} F(x)&=\int_{-\infty}^\infty \frac{1}{1+2\pi i\xi}\,e^{2\pi i\xi x}\,d\xi\\\\ &=\int_{-\infty}^\infty \frac{1-2\pi i \xi}{1+4\pi^2 \xi^2}\,e^{2\pi i\xi x}\,d\xi\\\\ &=\int_{-\infty}^\infty \frac{\cos(2\pi \xi x)}{1+4\pi^2\xi^2}\,d\xi+2\pi \int_{-\infty}^\infty \frac{\xi\sin(2\pi \xi x)}{1+4\pi^2\xi^2}\,d\xi \\\\ &=\frac1{2\pi }\int_{-\infty}^\infty \frac{\cos(\xi x)}{1+\xi^2}\,d\xi+\frac1{2\pi}\int_{-\infty}^\infty \frac{\xi \sin(\xi x)}{1+\xi ^2}\,d\xi\\\\ &=\frac1{2\pi }\int_{-\infty}^\infty \frac{\cos(\xi x)}{1+\xi^2}\,d\xi+\frac1{2\pi}\int_{-\infty}^\infty \frac{((\xi^2+1)-1) \sin(\xi x)}{\xi(1+\xi ^2)}\,d\xi\\\\ &=\frac1{2\pi }\int_{-\infty}^\infty \frac{\cos(\xi x)}{1+\xi^2}\,d\xi+\frac1{2\pi}\int_{-\infty}^\infty \frac{\sin(\xi x)}{\xi}\,d\xi-\frac1{2\pi}\int_{-\infty}^\infty \frac{ \sin(\xi x)}{\xi(1+\xi ^2)}\,d\xi\\\\ &=\frac1{2\pi }\int_{-\infty}^\infty \frac{\cos(\xi x)}{1+\xi^2}\,d\xi+\frac1{2}\text{sgn}(x)-\frac1{2\pi}\int_{-\infty}^\infty \frac{ \sin(\xi x)}{\xi(1+\xi ^2)}\,d\xi\\\\ \end{align}$$
Now, differentiation under the first integral is justified since $\int_{-\infty}^\infty \frac{\xi \sin(\xi x)}{1+\xi^2}\,d\xi$ converges uniformly for all $|x|>\delta>0$. Differentiation under the second integral is justified since the derivative of the integrand is absolutely integrable.
We split the procedure of differentiation into the two cases for which $x>0$ and $x<0$. In both cases, we find that
$$\begin{align} F'(x)&=-\frac1{2\pi}\int_{-\infty}^\infty \frac{\xi \sin(\xi x)}{1+\xi^2}\,d\xi-\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\cos(\xi x)}{1+\xi ^2}\,d\xi\\\\ &=-F(x) \end{align}$$
Therefore, $F(x)=C^{+}e^{-x}$, for $x>0$ and $F(x)=C^- e^{-x}$ for $x<0$.
Now, note that as $x\to 0^+$, $F(x)\to 1$ while as $x\to 0^-$, $F(x)\to 0$. We conclude that
$$F(x)=e^{-x}H(x)$$
where $H(x)$ is the heaviside function. And we are done!