# 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

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
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)