Integral of a rational function with varying exponents

Question

Find the exact value of the following integral, where $r,s \in \mathbb{R} $ and $ 0<r<s$. $$\int_{0}^{\infty}\frac{x^{r-1}}{1+x^s}dx$$

I have looked 3 cases:
Case 1: $0<r,s<1$
Case 2: $0<r<1<s$
Case 3: $1<r<s$
Under all 3 cases, the integral is convergent from 0 to $\infty$. The area under the curve varies with r and s. How can I find the "exact value" of the integral?

Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. — José Carlos Santos, Mar 03 '22 at 03:07
See this case for $s=2$. The integral may be a lerch transcendent — Тyma Gaidash, Mar 03 '22 at 05:19

score 1 · Accepted Answer · answered Mar 03 '22 at 06:43

Hint

$$I=\int\frac{x^{r-1}}{1+x^s}dx$$

$$x^s=t \implies x=t^{\frac{1}{s}}\implies dx=\frac 1 s t^{\frac{1}{s}-1} \,dt$$ $$I=\frac{1}{s}\int \frac {t^a } {1+t}\,dt \qquad \text{where} \qquad a=\frac r s-1 <0$$

Under the conditions $r,s \in \mathbb{R}$ and $0 <r <s$, the definite integral $$J=\int_0^\infty \frac {t^a } {1+t}\,dt$$ is pretty well known.

Integral of a rational function with varying exponents

1 Answers1