How to evaluate $ \int_{0}^{\infty} xe^{-ix(a-b)} d x $ I would like to solve the following integral:
$$
\int_{0}^{\infty} xe^{-ix(a-b)} d x
$$
a and b are real numbers.
Edit: the result should be $$\frac{1}{(a-b)^2}$$
I just found it on internet, but I still don't know how to demostrate it.
 A: The integral isn't convergent but can be reinterpreted as a Fourier transform of a distribution.
Using the "non-unitary, angular frequency" variant of the Fourier transform,
$$
\mathcal{F}\{ f(x) \} = \int_{-\infty}^{\infty} f(x) \, e^{-i\nu x} \, dx,
$$
we can interpret the given integral as $\mathcal{F}\{ x\, u(x) \},$ where $u(x)$ is the Heaviside step function, then evaluated for $\nu=a-b.$
Using properties 107 and 313 we get
$$
\mathcal{F}\{ x \, u(x) \}
= i\frac{d}{d\nu} \mathcal{F}\{ u(x) \}
= i\pi \frac{d}{d\nu} \left( \frac{1}{i\pi\nu} + \delta(\nu) \right)
= i\pi \left( -\frac{1}{i\pi\nu^2} + \delta'(\nu) \right)
= -\frac{1}{\nu^2} + i\pi\delta'(\nu).
$$
Thus,
$$
\int_{0}^{\infty} x e^{-i(a-b)x} \, dx
= -\frac{1}{(a-b)^2} + i\pi\delta'(a-b).
$$
With the distributional meaning of these terms it's not necessary to put a condition $a\neq b,$ but for a classical meaning we require $a\neq b$ and then the second term vanishes and we are left with
$$
\int_{0}^{\infty} x e^{-i(a-b)x} \, dx
= -\frac{1}{(a-b)^2}.
$$
A: Start by writing the integral in the following way:
\begin{eqnarray}
I(k)=\int_{0}^{\infty}dx xe^{-ikx}, \hspace{10pt}k=a-b.
\end{eqnarray}
The integrand can be rewritten as:
\begin{eqnarray}
I(k)=i\int_{0}^{\infty}dx\frac{d}{dk}\left(e^{-ikx}\right)
\end{eqnarray}
Now consider performing the indefinite integral in $k$ and switch the order of integration on the RHS. This gives:
\begin{eqnarray}
\int I(k)dk = i\int_{0}^{\infty}dx\int dk\frac{d(e^{-ikx})}{dk}=i\int_{0}^{\infty}dxe^{-ikx} +C
\end{eqnarray}
where $C$ is an arbitrary constant. Now, the above integral is not convergent; you might use a convergence parameter $\epsilon>0$ to write:
\begin{eqnarray}
\lim_{\epsilon\to0^{+}}\int_{0}^{+\infty}dx e^{-ikx-\epsilon x}=\frac{e^{-ikx-\epsilon x}}{-ik-\epsilon}\bigg|_{0}^{\infty}=+\frac{1}{ik+\epsilon}
\end{eqnarray}
Multiplying by the prefactor $i$ we get:
\begin{eqnarray}
\lim_{\epsilon\to0^{+}}\frac{i}{ik+\epsilon}=\frac{1}{k}
\end{eqnarray}
So in total we have:
\begin{eqnarray}
\int dk I(k)=C+\frac{1}{k}
\end{eqnarray}
The function $I(k)$ that provides this indefinite integral is:
\begin{eqnarray}
I(k)=\frac{(-1)}{k^{2}}=\frac{(-1)}{(a-b)^{2}}
\end{eqnarray}
so the integral is evaluated.
