Can a general formula be developed for these integrals? Can the following integral $I$ be written wrt $n$?
$$I=\displaystyle\int_{0}^{\infty} \dfrac{dx}{\sum_{k=0}^{n} x^k}$$
I found the values for $n=1,2,3$ 
but can we generalize it for an arbitrary $n$? 
 A: Because
$$\sum_{k=0}^{n} x^k=\frac{1-x^{n+1}}{1-x}$$
$$I_{n+1}=\displaystyle\int_{0}^{\infty} \dfrac{dx}{\sum_{k=0}^{n} x^k}=\int_0^{\infty}\frac{1-x}{1-x^{n+1}}dx$$
Mathematica 7.0 produces (after simplification) ($n>2$):
$$I_n=\cos\left(\frac{\pi}{n}\right)\Gamma\left(1 - \frac1n\right)
     \Gamma\left(1 + \frac1n\right) - 
   (1/2)\cos\left(\frac{2\pi}{n}\right)\Gamma\left(1 - \frac2n\right)
     \Gamma\left(1 + \frac2n\right)$$
EDIT
This expression can be further simplified as pointed out by David H. and Lucian in the comments.
The Euler's reflection formula for the Γ function is given by:
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}......(1)$$
We also have
$$\Gamma(1+z)=z\Gamma(z)......(2)$$
From (1) and (2) we obtain
$$\Gamma(1+z)\Gamma(1-z)=\frac{\pi z}{\sin(\pi z)}......(3)$$
Substitution of (3) into the expression for $I_n$ leads to the neat expression:
$$I_n=\cos\left(\frac{\pi}{n}\right)\frac{\pi/n}{\sin(\pi/n)} - 
   (1/2)\cos\left(\frac{2\pi}{n}\right)\frac{2\pi/n}{\sin(2\pi/n)}=\frac{\pi}{n}\csc{\frac{2\pi}{n}}$$
