Evaluate an integral with Euler Beta function

Question

I'm trying to express the integral $$ \int_1^\infty (x^r - 1)^p x^{-q} \,dx $$ in forms of the Euler Beta function where $pr + 1 < q$ s o the integral converges. I tried the substitution $u = x^r - 1$ but I don't know how to proceed.

Answer 1

HINTS:

$(1)$ Enforce the substitution $x\to 1/x$ so that the new limits are from $0$ to $1$.

$(2)$ Next, write $(x^{-r}-1)^p=x^{-rp}(1-x^r)^p$.

$(3)$ Enforce the substitution $x\to x^{1/r}$

Can you finish now?

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Following the hints we have$$\begin{align}\int_1^\infty (x^r-1)^p x^{-q}\,dx&=\int_0^1 (x^{-r}-1)^p x^{q-2}\,dx\\\\&=\int_0^1 (1-x^r)^p x^{q-rp-2}\,dx\\\\&=\frac1r \int_0^1 (1-x)^{(p+1)-1} x^{\frac{q-rp-1}{r}-1}\,dx\\\\&=\frac1r B\left(p+1,\frac{q-rp-1}{r}\right)\end{align}$$

Answer 2

Upon your substitution we have $du=rx^{r-1}dx$ so $\frac{1}{r}x^{1-r}du=dx$. And $\frac{1}{r}(u+1)^{1/r(1-r)} du$:

$$\frac{1}{r} \int_{0}^{\infty} u^p (u+1)^{(-1/r)q}(u+1)^{1/r(1-r)} du$$

$$\frac{1}{r} \int_{0}^{\infty} \frac{u^p}{(1+u)^{1/r(r-1+q)}} du$$

Now note one representation ,

$$B(x,y)=\int_{0}^{\infty} \frac{t^{x-1}}{(1+t)^{x+y}} dt$$

To prove this let $u=\sqrt{\frac{t}{t-1}}$ and continue from there.