Show that $\int_0^1\frac{\sqrt{\sin x}}{x}dx$ converges

Question

Show that $\displaystyle\int_0^1\frac{\sqrt{\sin x}}{x}dx$ converges

I couldn't find an anti-derivative and all convergence tests I tried don't seem to work here. Hints are welcome.

Answer 1

Notice that $ x\mapsto \frac{\sqrt{\sin{x}}}{x} $ is continuous and positive on $ \left(0,1\right] $ so we can use the comparaison test : $$ \left(\forall x\in\left[0,1\right]\right),\ \sin{x}\leq x $$

Thus : $$ \left(\forall x\in\left(0,1\right]\right),\ \frac{\sqrt{\sin{x}}}{x}\leq\frac{1}{\sqrt{x}} $$

Since $ \int_{0}^{1}{\frac{\mathrm{d}x}{\sqrt{x}}} $ converges, $ \int_{0}^{1}{\frac{\sqrt{\sin{x}}}{x}\,\mathrm{d}x} $ converges.

Answer 2

Use the Taylor expansion of $\sin(x)$ and then the binomial expansion to get $$\sqrt{\sin(x)}=x^{1/2}-\frac{x^{5/2}}{12}+\frac{x^{9/2}}{1440}+O\left(x^{11/2}\right)$$ $$\frac{\sqrt{\sin (x)}}{x}=x^{-1/2}-\frac{x^{3/2}}{12}+\frac{x^{7/2}}{1440}+O\left(x^{9/2}\right)$$ $$\int \frac{\sqrt{\sin (x)}}{x}\, dx=2 x^{1/2}-\frac{x^{5/2}}{30}+\frac{x^{9/2}}{6480}+O\left(x^{11/2}\right)$$ $$\int_0^1 \frac{\sqrt{\sin (x)}}{x}\, dx \sim \frac{2549}{1296} \approx 1.966821$$ while numerical integration would give $1.966814$