Derivate $F$ where $F$ is defined by a Lebesgue integral.

Question

Calculate $F'(x)$ where $F$ is defined by Lebesgue integral $$F(x)= \int_0^1 \frac{sin(xt)}{1+t}dt$$

It is my first time to calculate these kind of derivate, and I don't find exmaples to do it. So what I try is just do it as a simple integral but that integral doesn't have an antiderivate known. So could you help me with an example or hint?

Answer 1

Hint

First, it's a Riemann integral (Riemann and Lebesgue integral coincides here). Moreover, $(x,t)\mapsto \frac{\sin(xt)}{1+t}$ is $\mathcal C^\infty (\mathbb R\times [0,1])$. So by a famous theorem, $$\frac{d}{dx}\int_0^1\frac{\sin(xt)}{1+t}dt=\int_0^1\frac{\partial }{\partial x}\frac{\sin(xt)}{1+t}dt.$$ I let you conclude.

Answer 2

This problem can be turned into a problem of interchanging orders of integration instead, which is fairly common for such problems: $$ F(x) = \int_0^1\frac{\sin(xt)}{1+t}dt= \int_0^1 \frac{\sin(ut)|_{u=0}^{x}}{1+t}dt \\ = \int_0^1 \frac{\int_0^xt\cos(ut)du}{1+t}dt \\ T= \int_0^x\left(\int_0^1\frac{t\cos(ut)}{1+t}dt\right)du $$ The inner integral is a continuous function of $u$. Therefore, by the Fundamental Theorem of Calculus $$ F'(x) = \int_0^1\frac{t\cos(xt)}{1+t}dt $$