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

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?

• The limits of the integration are a function of $x$, namely $a(x) = \infty$ and $b(x) = -\infty$, thus the functions are just constant functions. – Hetebrij Aug 13 '16 at 7:36
• There is no real (or complex) value $x$ for which this integral makes sense, neither does the limit $\int_{-N}^N e^{ixt}\>dt$ when $N\to\infty$ exist. – Christian Blatter Aug 13 '16 at 8:00
• If you want to know when you can differentiate under the integral sign, see my answer here (make sure you read through the required three conditions in the OPs post). – Mattos Aug 13 '16 at 9:08
• @Mattos I don't think these conditions would apply since I have an $i$ in the exponent. Can these conditions be modified? – cap Aug 13 '16 at 21:22
• @cap As long as your integrand satisfies the three conditions, then you can differentiate under the integral. – Mattos Aug 14 '16 at 0:09

## 2 Answers

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 $f$.

• Your $e$ should have a negative in the exponent in this case. – cap Aug 13 '16 at 21:07
• Can you prove this or cite criteria needed to differentiate under the integral for such functions? – cap Aug 13 '16 at 21:24
• @cap: In that case, one would have to define precisely what version of the Fourier transform is used. And yes, that is easy to prove for test functions with compact support and slightly more involved for tempered test functions. – LutzL Aug 14 '16 at 0:01

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)

• "In general" well no not in general there needs to be some other conditions. – user223391 Aug 13 '16 at 21:04