$$\begin{align} & \int_{0}^{+\infty }{\frac{\sin px}{1+{{\text{e}}^{qx}}}}\text{d}x ,\ \ p,\ q>0\\ \\ \\ & \int_{0}^{+\infty }{{{\left( \frac{\sin x}{x} \right)}^{n}}\text{d}x} \\ \end{align}$$
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Are you stuck on both? What have you tried? – MITjanitor Jan 25 '13 at 15:51
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What is your question? – JRN Feb 20 '15 at 15:47
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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
$$\int_0^{\infty} dx \: \frac{\sin{p x}}{1+e^{q x}} = \frac{1}{2 p} - \frac{\pi}{2 q} \mathrm{csch}{\pi \frac{p}{q}} $$
Ron Gordon
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It is funny for me that you always write $dx$ first in any integral. Do you have a certain reason for that. :-) +1 – Mikasa Jan 25 '13 at 16:31
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It's because of my background in optics, where we work mainly with multiple integrals. It's easier to keep track of what integral goes with which variable writing them this way. It just looks cleaner to me. I know it annoys a lot of people here; my apologies if it does. – Ron Gordon Jan 25 '13 at 16:32
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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: $$\Big|\frac{\sin px}{1+e^{qx}}\Big|<\text{e}^{-qx}$$
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@amWhy: Ohhh. Thanks. I didn't know that. Thanks for the edit and the time. – Mikasa Jan 27 '13 at 15:20