Differentiating $\int_{-\infty}^{\infty}e^{ixt}dt$

Question

Why is the derivative of $F(x)=\int_{-\infty}^{\infty}e^{ixt}dt$ (with respect to $x$) equal to $i\int_{-\infty}^{\infty} te^{itx}dt$?

If I ignore the integral sign, I see that $\frac{d}{dx}e^{itx}=e^{itx}it$ by the chain rule, but I don't see why I am allowed to disregard the integral sign. I don't think the fundamental theorem of calculus applies since due to the limits of integration not being functions of $x$.

Edited What conditions have to be checked in order to differentiate this type of function (with imaginary number in integrand) under the integral sign?

Answer 1

In general, $$\frac{d}{dx}\int_a^b f(x, t) dt = \int_a^b \frac{\partial}{\partial x}f(x, t) dt$$ So here, we have $$F'(x) = \int_{-\infty}^\infty \frac{\partial }{\partial x}e^{ixt} dt = i\int_{-\infty}^\infty te^{ixt} dt$$ (you can't factor the $t$ out of the integral as you seem to have done)

Answer 2

Neither the integral nor the derivative exist in the classical sense. As a somewhat abusive notation for distributions, this is the relation between the Fourier transform and derivatives, if $$ F(x)=\int_{-\infty}^\infty e^{ixt}f(t)dt $$ then $$ F'(x)=\int_{-\infty}^\infty ite^{ixt}f(t)dt $$ for all fast falling test functions $f$.