Interchanging integral and sum for $\sum_{n=1}^{\infty} \int_0^{\infty} \frac{\sin(nx)}{n!x}dx$

Question

Good afternoon. I'm struggling on a problem, I would like to show that :

$$\int_0^{\infty} \sum_{n=1}^{\infty} \frac{\sin(nx)}{n!x}dx = \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\sin(nx)}{n!x}dx = \frac{\pi}{2}(e-1)$$

The second equality isn't the problem, I'm trying to justify the first one. What I tried for the moment :

ATTEMPT 1 : Let $f_n(x) := \frac{\sin(nx)}{n!x}$. I tried to apply Fubini's theorem by showing : $$\int_0^{\infty} \sum_{n=1}^{\infty} |f_n(x)| < \infty$$

It didn't work, even by cutting the integral (from $0$ to $1$ and from $1$ to $\infty$).

ATTEMPT 2 : By uniform convergence of $(f_n(k))_n$ we have :

$$\int_0^{b} \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} \int_0^{b} f_n(x)$$

Thus : $$\int_0^{\infty} \sum_{n=1}^{\infty} f_n(x) = \lim\limits_{b\to \infty} \int_0^{b} \sum_{n=1}^{\infty} f_n(x) = \lim\limits_{b\to \infty} \sum_{n=1}^{\infty} \int_0^{b} f_n(x)$$ $$= \sum_{n=1}^{\infty} \lim\limits_{b\to \infty} \int_0^{b} f_n(x) = \sum_{n=1}^{\infty} \int_0^{\infty} f_n(x)$$

But It looks like a scam, I'm not convinced at all by the penultimate equality.

Any idea ?

Answer 1

Not an answer. Only prove the uniform convergence.

Define the partial sum:

$$S_n(x)=\sum_{k=1}^{n} \frac{\sin(kx)}{k!x}$$

Let $E=(0,\infty)$, we want to show:

$$S_n(x) \xrightarrow{\text{Uniform on}~E} S(x)$$

We have $$\begin{align}\sup_{x\in E}\left|S_{n}(x)-S(x) \right|&=\sup_{x\in E}\left|\sum_{k=n+1}^{\infty} \frac{\sin(kx)}{k!x} \right|\\ \\ &\le\sup_{x\in E}\sum_{k=n+1}^{\infty} \left|\frac{\sin(kx)}{k!x} \right|\\ \\ &=\sup_{x\in E}\sum_{k=n+1}^{\infty} \frac{1}{(k-1)!}\frac{|\sin(kx)|}{|kx|}\\ \\ &\le\sup_{x\in E}\sum_{k=n+1}^{\infty} \frac{1}{(k-1)!}\\ \\ &=\sum_{k=n+1}^{\infty} \frac{1}{(k-1)!} \longrightarrow 0 \end{align}$$