could you please help me. And let me notice how bad I am please.
Problem: Calculate the derivative $F^\prime(x)$ of the function $F$ defined by the Lebesgue integral
$$F(X) = \int_{0}^{1} \dfrac{\sin {xt}}{1+t} \ dt , x\in [0,1]$$
My attempt:
Since the function $\dfrac{\sin {xt}}{1 + t}$ is a bounded function on $[0, 1]$, then the integral (Riemann usual) $\int_{0}^{1} \dfrac{\sin {xt}}{1+t} \ dt $converges, which implies that the integral of Lebesgue $\int_{0}^{1} \dfrac{\sin {xt}}{1 + t} \ dt $ exists, so $f$ is summable.
Then using Theorem 9.6 which tells us that whatever the summable function $f$ in the segment $[a, b] \subset \mathbb{R}$, in almost all points the equality
$$\dfrac{\mathrm{d} }{\mathrm{d} x}\int_{a}^{x}f(t) \ dt = f(x)$$
Then the $F^\prime (x)$ is equal to.
$$F^\prime(x) = \dfrac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{1}\frac{\sin {xt}}{1+t} \ dt = \dfrac{\sin {x(1)}}{1+(1)} = \dfrac{\sin {x}}{2}$$
Many thanks
No, its wrong. The theorem you quote is for functions $f$ that do not depend on $x$, and then the $x$ is in the integral limits. Neither hold in this case. Furthermore, $F(x)$ is an odd function; hence $F'$ must be an even function, which $(\sin x)/2$ is not.
You should check with appropriate limit theorems that you are allowed to differentiate under the integral sign, giving $$ F'(x) = \int_0^1 \frac{\cos(xt)tdt}{1+t}.$$ I don't know what sort of answer you're looking for, but Wolfram tells me that there is an expression in terms of special functions:
$$\int_{0}^{1} \frac{\cos (x t) t}{1+t} d t=\operatorname{Ci}(x) \cos (x)-\operatorname{Ci}(2 x) \cos (x)+\operatorname{Si}(x) \sin (x)+\left(\frac{1}{x}-\operatorname{Si}(2 x)\right) \sin (x).$$