computing $\int_{-\infty}^{\infty} x e^{itx} dx$ using Cauchy Theorem and gamma function rapresentation

Question

I want to solve the above integral with $t \in \mathbb{R}$; using a software (Wolfram) to verify the result I obtain the following result:
$I(t) \equiv \int_{-\infty}^{\infty} x e^{itx} dx = -i \frac{d}{dt}\int_{-\infty}^{\infty} e^{itx} dx = -i2\pi \frac{d \delta(t)}{dt}$
To solve it I start rewritting $I(t)$ as:
$\int_{-\infty}^{\infty} x e^{itx} dx = \int_{-\infty}^{0} x e^{itx} dx + \int_{0}^{\infty} x e^{itx} dx = -\int_{0}^{\infty} w e^{-itw} dw + \int_{0}^{\infty} y e^{ity} dy $
I tried to solve both the last integrals using Cauchy's integral theorem; In my case I have (following the notation of the last link) $\Re(\beta) = 0$, but this shouldn't be a problem, In fact I can write $t = lim_{h \rightarrow 0} t \pm ih$ and interchange the limit with the integral (using DCT).
With this approach I obtain the following result:
$I(t) = -\frac{\Gamma(2)}{(it)^2} + \frac{\Gamma(2)}{(-it)^2}$
which is 0 for all $t$ except for $t=0$ where I have an indeterminate form $-\infty + \infty$.
Is it correct this solution?

Answer 1

No.

$$I(t) = \int_{-\infty}^{\infty} x e^{itx} dx$$ diverges for all $t$. It converges only in the sense of distributions.

$$I_N(t)=\int_{-N}^N x e^{itx} dx=-2i \frac{d}{dt}\frac{\sin(Nt)}{t}$$ $$h(t)=\int_{-\infty}^t 2\frac{\sin(u)}{u}du, \qquad h(Nt)\to 2\pi 1_{t> 0}$$ For all $\phi \in C^\infty_c$ $$\lim_{N\to \infty} \int_{-\infty}^\infty I_N(t)\phi(t)dt=\lim_{N\to \infty} \int_{-\infty}^\infty i\, h(Nt)\phi''(t)dt= \int_{-\infty}^\infty 2i\pi 1_{t>0}\phi''(t)dt=2i\pi\phi'(0)$$ This is the definition of $\int_{-\infty}^{\infty} x e^{itx} dx$ converges to $-2i\pi\delta'(t)$ in the sense of distributions.

To use the Gamma function you need to do the same kind of idea, saying that $\lim_{a\to 0}\int_{-\infty}^{\infty} x e^{itx}e^{-a|x|} dx$ converges to $\int_{-\infty}^{\infty} x e^{itx} dx$ in the sense of distributions.

You'll get (something like) $\lim_{a\to 0} \frac1{(it+a)^2}-\frac1{(it-a)^2}$, the second derivative of $\lim_{a\to 0} \log(it-a)-\log(it+a)=C\ sign(t)$