What went wrong with integrating $\frac {x^a}{1-x^b}$?

Question

Question: Integrate$$I=\int\limits_0^{\infty}dx\,\frac {x^a}{1-x^b}$$

I started off with the substitution $u=x^b$ so $du=bx^{b-1}$. Therefore, $I$ is equal as$$I=\frac 1b\int\limits_0^{\infty}du\, (1-u)^{-1}u^{(a-b+1)/b}$$Next I made the substitution $v=(1-u)^{-1}$ so $dv=-(1-u)^{-2}\ du$. Now, $I$ is equal to$$I=\frac 1b\int\limits_0^{1}dv\, v^{-1}\left(\frac {v-1}{v}\right)^{(a-b+1)/b}=\frac 1b(-1)^{(a-b+1)/b}\int\limits_0^1dv\, (1-v)^{(a-b+1)/b}v^{-(a-b+1)/b-1}$$However, evaluating the beta function on the right-hand side gives$$I=\frac {\pi}b\color{red}{(-1)^{(a-b+1)/b}\csc}\left(\frac {\pi(a+1)}b\right)$$Which is definitely wrong because the answer is supposed to be$$I=\frac {\pi}b\cot\left(\frac {\pi(a+1)}b\right)$$Can anybody help me figure out what I did wrong in my work?

Answer 1

The only mistake I see is in the first substitution, the exponent of $u$ should be $a/b-b+1$.

Answer 2

Your first substitution is fine (but write $dx$ and $du$ at the end of integral): $$x^b=u, x=u^{\frac{1}{b}},x^a=u^{\frac{a}{b}},\\ bx^{b-1}dx=du \to b(u^{\frac{1}{b}})^{b-1}dx=du \to dx=\frac{1}{b}u^{\frac{1}{b}-1}du$$ $$\int_0^{\infty} \frac{x^a}{1-x^b}dx=\int_0^{\infty}\frac{u^{\frac{a}{b}}}{1-u}\cdot \frac{1}{b}u^{\frac{1}{b}-1}du=\frac{1}{b}\int_0^{\infty} (1-u)^{-1}u^{\frac{a}{b}+\frac{1}{b}-1}du=\\ \frac{1}{b}\int_0^{\infty} (1-u)^{-1}u^{\frac{a-b+1}{b}}du.$$ Your second substitution must be: $$v=(1-u)^{-1} \to dv=(1-u)^{-2}du, \\ v(0)=1,v(\infty)=0.$$ note that there is no minus sign, because of the chain rule.

Next step must be: $$I=\frac 1b\int\limits_1^{0} v\cdot \left(\frac {v-1}{v}\right)^{(a-b+1)/b}dv=\frac 1b(-1)^{(a+1)/b}\int\limits_0^1 (1-v)^{(a-b+1)/b}v^{-(a-b+1)/b+1}dv.$$