Derivative of an improper integral function

Question

Let define, for $t>0$, the function

$$f(t)=\int_0^{+\infty}\frac{\sin x}{x}e^{-tx}dx$$

How to compute $\frac{d}{dt}f(t)$?

My idea is to calculate

$$\frac{d}{dt} \int_\epsilon^{N}\frac{\sin x}{x}e^{-tx}dx$$ and then take the limits for $N \rightarrow +\infty$ and $\epsilon \rightarrow 0$, does it work?

Thank you!

Answer 1

$$\frac{d}{dt}\int_{0}^{+\infty}\frac{\sin x}{x}e^{-tx}\,dx \color{red}{=}\int_{0}^{+\infty}\frac{d}{dt}\frac{\sin x}{x}e^{-tx}\,dx = -\int_{0}^{+\infty}\sin(x)e^{-tx}\,dx = -\frac{1}{t^2+1}$$ Where $\color{red}{=}$ holds by the dominated convergence theorem, since for any $t>0$ $$\left|\int_{0}^{+\infty}e^{-tx}\sin x\,dx\right|\leq \int_{0}^{+\infty}|\sin x|e^{-tx}\,dx \leq \int_{0}^{+\infty}e^{-tx}\,dx = \frac{1}{t}.$$

Answer 2

By using the definition you get $f'(t)=-\int_0^{+\infty}{\sin x}e^{-tx}dx$ but you have to prove this integral is convergent.

Answer 3

Notice, you can use the Laplace transform:

$$\text{F}(\text{s})=\mathcal{L}_x\left[\frac{\sin(x)}{x}\right]_{(\text{s})}=\int_0^\infty\frac{\sin(x)}{x}\cdot e^{-\text{s}x}\space\text{d}x=\arctan\left(\frac{1}{\text{s}}\right)$$

When $\Re[\text{s}]>0$

So, ifwe want to find $\text{F}'(s)$:

$$\text{F}'(s)=\frac{\text{d}}{\text{d}\text{s}}\left(\arctan\left(\frac{1}{\text{s}}\right)\right)=-\frac{1}{1+\text{s}^2}$$