Help me solve an equation!

[hamidr_karami]

solve

hi all
please solve this

/ inf
|
| (x^(u-1))*sin(x)dx = ???
|
/ 0

bye

 

[hamidr_karami]

Re: solve

hi all

u is a constant.

 

[steve10]

Re: solve

You got it.

 

[hamidr_karami]

Re: solve

hi steve10
thank you for it / inf z^(-u) Pi
i know | -------- dz =------------
/ 0 1+z^2 2*sin((1-u)*pi/2)

but i don't infer it.
please infer it for me.

 

[steve10]

Re: solve

hi steve10
thank you for it / inf z^(-u) Pi
i know | -------- dz =------------
/ 0 1+z^2 2*sin((1-u)*pi/2)

but i don't infer it.
please infer it for me.

The above is the direct quote from your last message. Can you read it? Did you expect I could read it? Are you serious about your question, or are you just fooling around here?

In any case, I'll still make a guess and try to answer it now, but not later.

Try to calculate Integrate[z^(-u)/(1+z^2),{z,0,Infinity}]? There are a couple ways to resolve this integral. One of them is through infinte product of sin[x]. The process is quite long and, therfore, we'll forget it. The easiest way (to me) would be using the Residue Theorem. Here are the steps:

1. Set f(z)= z^(-u)/(1+z^2) (notice that, -1<u<1) and treat it as a function in the complex plan;
2. The two branches of f(z) are 0 and Infinity. Make a cut to connect these two points through the positive real axis;
3. Notice that there are two poles of f(z), -i and i;
4. Make a contour which consists of four pieces: A big circle of radius M centered at the z=0, a small circle of radius m centered at z=0, two straight lines along the positive real axis with arg = 0 and 2*Pi respectively and with the radii from m to M. Later you need to let M->Infinity and m->0.

The rest only involves some algebraic operations. Good luck.

 

[hamidr_karami]

Re: solve

Hi Steve excuse me
And thank you for it.
Regards