How to solve this question related to Definite Integration How to solve this question please help $$\int_0^\infty \frac{x^{n-1}}{1+x}dx$$
 A: A related problem. Make the change of variables $ x=\frac{1}{u}-1 $ and use the beta function.
$$ \int_0^\infty \frac{x^{n-1}}{1+x}dx = \int_0^1 u^{-n}(1-u)^{n-1} du = \Gamma(n)\Gamma(1-n),$$
Note: the integral exists for $0<n<1.$ 
A: If I were you, I would try to prove that the integral doesn't exist. 
EDIT: As Mhenni notes, it exists for $0\lt n\lt1$. For $n=1/2$, it's not so hard to evaluate. The substitution $u=\sqrt x$, $x=u^2$, $dx=2u\,du$ turns it into an integral involving $1/(1+u^2)$, which has antiderivative $\arctan u$. For other rational values of $n$, a similar substitution will result in a rational function which can, in principle, be integrated by partial fractions (in practice, it might not be so much fun). For general $n$, see Mhenni's answer. You may have to read up on the Gamma function to understand Mhenni's answer; Wikipedia will get you started. 
A: The integral does not exist when $n\geqslant1$ or when $n\leqslant0$, as simple equivalents of the function integrated when $x\to0$ or when $x\to+\infty$ show. For every $s$ in $(0,1)$,
$$\int_0^{+\infty}\frac{x^{s-1}}{1+x}\mathrm dx=\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}.
$$
The first equality is already mentioned on this page. The second is called Euler's reflection formula. A short proof using Weierstrass products is here. A longer proof, but entirely elementary, is there.
