Question

integrate..............

Answer 1

\[\int\limits_{0}^{\infty}\frac{ x^{n-1} }{ 1+x }dx\]

Answer 2

@satellite73

Answer 3

Ummm, did you try integration by parts?

Answer 4

\[\int_0^\infty\frac{x^{n-1}}{1+x}dx\\
\lim_{b\to\infty}\int_0^b\frac{x^{n-1}}{1+x}dx\]
\[u=1+x\iff x=u-1\\
du=dx\]
\[\lim_{b\to\infty}\int_0^b\frac{(u-1)^{n-1}}{u}du\]
Expanding the numerator is the toughest part. Are you familiar with the Binomial Theorem? http://en.wikipedia.org/wiki/Binomial_theorem

Answer 5

yes yes iam familiar with it could u plz elaborate henceforth.............

Answer 6

\[ \large
(a+b)^m = \sum_{k=0}^{m}\binom{m}{k}a^{m-k}b^k
\]Letting \(a=u,b=-1,m=n-1\)\[ \large
(u-1)^{n-1} = \sum_{k=0}^{n-1}\binom{n-1}{k}u^{n-1-k}(-1)^k
\]

Answer 7

but still i did not get it after that ................. the answer is npi/sin npi

Answer 8

\[ \Large
\sin(x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}
\]

Answer 9

how come pi come there

Answer 10

\[ \Large
\pi = 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}
\]

Answer 11

ok..........thanks...............but how come pi equal to that

Answer 12

Because \(\pi\) is a transcendental number.