integrate

(x^(n-1))/(1+x)

over the limit from 0 to infinty. It is given 0

Question

integrate  
                       
                          (x^(n-1))/(1+x) 

over the limit from 0 to infinty. It is given 0<n<1.

anonymous · Answer

Integrate with respect to x.
The answer is = pi/sin(n*pi)

anonymous · Answer

It can be done using Cauchy residue theorem. But can it done in simpler ways (substitution etc...)

mathmale · Answer

You might want to experiment with re-writing the integrand in various ways until you find a form that you consider to be easy to integrate.  I am not familiar with the Cauchy residue theorem, and would thus recommend that you do an Internet search for that theorem and for examples of solved problems.

If 0<n<1, you might simplify this problem temporarily by letting n = 1/2 and attempting to integrate the resulting expression.

You could let u = x + 1 be a possible substitution; then du = dx.  Letting n = 1/2, where does this substitution take you?

kainui · Answer

I tried writing it as $$\int\limits_0^\infty \frac{x^ndx}{x(x+1)}$$ and tried partial fraction decomposition... lol $$\frac{x^n}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$$This gave me $$x^n=A(x+1)+Bx$$ since x can be anything, I set it equal to zero to get 0=A and then I set it to 1 to get B=1... $$\frac{x^n}{x(x+1)}=\frac{1}{x+1}$$ There's no way that's right though... Is it? lol

anonymous · Answer

we compare coeff of various x terms so that it remains true for all x.

kainui · Answer

I don't know, but I like the question.

anonymous · Answer

i think it is kind of special integral that cannot be solved by normal methods.

mathmale · Answer

I appreciate your intelligent and thoughtful approach to this question, as well as your willingness to consider and actually try out various possible approaches.

Would it help you to review that Cauchy residue theorem and examples of solved problems and then try to tackle the integration problem that you've posted?

anonymous · Answer

will look into it. thanks anyways!

kainui · Answer

From what I know, the Cauchy residue theorem is about integrating on a closed path around points that aren't analytic in the complex plane. This looks nothing like that, so I'm a little bit confused as to how that's going to work. I hope you figure it out, I love seeing new integrals.

mathmale · Answer

@diwakar :  Can you support your earlier statement, "It can be done using Cauchy residue theorem?"  Kainui has cast doubt on the applicability of this theorem to the problem you've posted.