Some integral with sine $$\begin{align}
  & \int_{0}^{+\infty }{\frac{\sin px}{1+{{\text{e}}^{qx}}}}\text{d}x ,\ \ p,\ q>0\\  \\ \\
 & \int_{0}^{+\infty }{{{\left( \frac{\sin x}{x} \right)}^{n}}\text{d}x} \\ 
\end{align}$$
 A: I will sketch an outline of the first integral.  There are convergence questions, and a sum that evaluates into a special function for which we will need a derivation.
Write the integral as
$$\int_0^{\infty} dx \: \frac{\sin{p x} e^{-q x}}{1+e^{-q x}} $$
Taylor expand the denominator:
$$\int_0^{\infty} dx \: \sin{p x} \, e^{-q x} \sum_{n=0}^{\infty} (-1)^n e^{-n q x} $$
Reverse the order of sum and integral.  Again, the justification is not trivial (see, e.g., Abel's Theorem):
$$\begin{align} \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} dx \: \sin{p x} \, e^{-(n+1) q x} &= \Im[\sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} dx \: e^{-[(n+1) q -i p]x} \\ &= \Im{\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)q - i p}} \\ &= p \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2 q^2 + p^2} \\ &= \frac{1}{2 p} \left [ 1 - \frac{p^2}{q^2} \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2 + \frac{p^2}{q^2}} \right ] \end{align} $$
I will not provide a derivation of the following closed form yet (perhaps in an update):
$$\begin{align} \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2 + a^2} = \frac{\pi}{a} \mathrm{csch}{\pi a} & (a>0) \\ \end{align}$$
(I can say that one derivation uses residues in the complex plane.)  Using this result, we find that
$$\int_0^{\infty} dx \: \frac{\sin{p x}}{1+e^{q x}} = \frac{1}{2 p} - \frac{\pi}{2 q} \mathrm{csch}{\pi \frac{p}{q}} $$
A: The first one,as you see in @rlgordonma's answer is convergent. The integral $$\int_a^{\infty}\text{e}^{-qx}dx$$ convergent if $q>0$. Also we see that: $$\Big|\frac{\sin px}{1+e^{qx}}\Big|<\text{e}^{-qx}$$
