More Integral

[AvgStudent]

Let [imath]r,s \in \R[/imath] be such that [imath]0 < r < s[/imath]. Find the exact value of the integral
[math]\int_{0}^{\infty}\frac{x^{r-1}}{1+x^s}\,dx[/math]No clue....?‍

 

[Dr.Peterson]

Let [imath]r,s \in \R[/imath] be such that [imath]0 < r < s[/imath]. Find the exact value of the integral
[math]\int_{0}^{\infty}\frac{x^{r-1}}{1+x^s}\,dx[/math]No clue....?‍

If I had no clue (and at this point I am in that state, not having even tried), I'd start by trying out a simple case, say r=1 and s=2, just to see if that looks doable. (If the result were 0 or infinite, which it won't be, that might be a clue.)

Then I might try some slightly larger numbers, and be looking for the possibility of a reduction formula (that is, using smaller numbers to solve it for larger numbers). On the other hand, since r and s aren't necessarily integers, I might try r=1/2 and s=1, or something like that. I'd just hope that some substitution, or parts, or whatever, might work in a specific case, and be able to be generalized.

Again, that's a deliberately ignorant beginning! I have no idea yet what might work.

 

[Steven G]

Have no clue how to proceed? I 2nd that you should try specific examples to see what is going on.

 

[Steven G]

To simplify things maybe you should let m = r-1 and s = r+k = r-1 + k-1 = m + k-1 = m + n
That is, replace r-1 with m and s with m+n.

 

[Steven G]

Did you consider what the graph of the function looks like? For example, if the integrand is always (or eventually) larger than 1 (or any positive real number), then the integral will be infinity. Do you see why that is true?
You don't always need to solve an integral in order to know its value.

 

[AvgStudent]

So I've divided the problem into 3 cases:
Case 1: 0<r<s<1


Case 2: 0<r<1<s


Case 3: 1<r<s


They are convergent from 0 to infinity, however, I don't how to even generalize this. How would you find the exact value?

 

[AvgStudent]

After wrestling with this problem for a while, I think I have the answer.
[math]\int_{0}^{\infty}\frac{x^{r-1}}{1+x^s}\,dx\\ Let \quad x^s=t\implies x=t^{1/s} \implies dx=\frac{1}{s}t^{1/s-1}dt\\ \frac{1}{s}\int_{0}^{\infty}\frac{t^{1/s(r-1)}\cdot t^{1/s-1}}{1+t}dt= \frac{1}{s}\int_{0}^{\infty}\frac{t^{r/s-1}}{1+t}dt\\ Let\quad c=r/s-1\\ \frac{1}{s}\int_{0}^{\infty}\frac{t^{c}}{1+t}dt\\ \text{This is the well known integral for} \csc(x):\\ -\frac{\pi}{s}\csc\left(\frac{\pi c}{s}\right) = -\frac{\pi}{s}\csc\left(\frac{\pi(r-s) }{s^2}\right)[/math]Thoughts and/or any mistakes?

 

[Steven G]

I will not accept your result until YOU (not copy!) prove that identity that you said is well known.

 

[AvgStudent]

I will not accept your result until YOU (not copy!) prove that identity that you said is well known.

[math] \frac{1}{s}\int_{0}^{\infty}\frac{t^{r/s-1}}{1+t}dt\\ Let \quad w=\frac{1}{t+1}:\\ \frac{1}{s}\int_{0}^{1}\frac{1}{w}\left(\frac{1}{w}-1\right)^{r/s-1}\,dw \\ \frac{1}{s}\int_{0}^{1}w^{-\frac{r}{s}}(1-w)^{\frac{r}{s}-1}\, dw=\frac{1}{s}\beta\left(1-\frac{r}{s},\frac{r}{s}\right)\\ \text{where }\beta(x,y) \text{is the beta function.}\\ \text{By the Beta-Gamma relation:}\\ \frac{1}{s}\beta\left(1-\frac{r}{s},\frac{r}{s}\right)=\frac{1}{s} \Gamma\left(1-\frac{r}{s}\right)\Gamma\left(\frac{r}{s}\right)\\ \text{By Euler's Reflection Formula:}\\ \frac{1}{s}\Gamma\left(1-\frac{r}{s}\right)\Gamma\left(\frac{r}{s}\right)= -\frac{1}{s}\cdot \frac{r}{s}\Gamma\left(\frac{r}{s}\right)\Gamma\left(-\frac{r}{s}\right)=-\frac{1}{s}\cdot\frac{\pi}{\sin(\frac{\pi r}{s})}=-\frac{\pi}{s}\csc\left(\frac{\pi r}{s}\right)\\ \therefore \int_{0}^{\infty}\frac{x^{r-1}}{1+x^s}\,dx=-\frac{\pi}{s}\csc\left(\frac{\pi r}{s}\right); 0<r<s \land r, s \in \R [/math]