Convergence of improper integral $\int_0^1 \frac{x^\alpha}{x+x^2}dx$ for $\alpha>0$

Question

I'm having trouble showing the convergence of the integral in the title, for $\alpha >0$:

$$\int_0^1 \frac{x^\alpha}{x+x^2}dx $$ I tried using:

$$\int_0^1 \frac{x^\alpha}{x+x^2}dx\leq \int_0^1 \frac{x^\alpha}{2x^2}dx=\frac{1}{2}\int_0^1 x^{\alpha -2}dx$$ But this integral converges for $\alpha>1$, not for $0<\alpha\leq1$.

Answer 1

Let $\alpha=1$.

Then, $$ \int_0^1\frac{x}{x+x^2}dx=\int_0^1\frac{1}{1+x}dx=\left.\ln(1+x)\right|_0^1=\ln(2) $$

Let $\alpha<1$.

Then, $$ \int_0^1\frac{x^\alpha}{x+x^2}dx=\int_0^1\frac{1}{x^{1-\alpha}(1+x)}dx $$ Since $1\leq(1+x)\leq 2$, $$ \int_0^1\frac{1}{x^{1-\alpha}(1+x)}dx\leq\int_0^1x^{\alpha-1}dx, $$ which converges for $\alpha<1$.

Answer 2

Since

$$0 < \frac{x^{\alpha}}{x + x^2} = \frac{x^{\alpha - 1}}{1 + x} < x^{\alpha -1}, \quad x > 0$$

and $\int_0^1 x^{\alpha - 1}\, dx = \frac{1}{\alpha} < \infty$, by the comparison test, $\int_0^1 \frac{x^\alpha}{x + x^2}\, dx$ convergees.