an improper integral or establish its divergence by definition

Question

$$\int_{0}^{1} \frac{\arcsin (x)}{x^{2}}dx$$ I tried to do it by parts and I came to $$\frac{\ln(\sqrt{1-x^{2}}-1)-\ln(\sqrt{1-x^{2}}+1)}{2}\bigg|_{0}^{1}-\frac{\arcsin x}{x}\bigg|_{0}^{1}$$ and here is still the non-proper points. So this integral is divergent? Or I made a mistake somewhere?

Answer 1

Letting $x=\sin t$ you have:

$$\int_{\epsilon}^{\pi/2} \frac{t\cos t}{\sin^2 t}\,dt$$

Letting $u=t, dv=\frac{\cos t}{\sin^2t}\,dt$ then $v=\frac{-1}{\sin t}$ and you get that the integral is equal to:

$$\frac{-t}{\sin t}\bigg|_{\epsilon}^{\pi/2}+\int_{\epsilon}^{\pi/2}\frac{dt}{\sin t}$$

But $\int \csc t \,dt =-\ln\left|\cot t+\csc t\right|+C$ so we have:

$$\left(\frac{-t}{\sin t}-\ln|\cot(t/2)|\right)\bigg|_{\epsilon}^{\pi/2}$$

So for $\epsilon>0:$ the integral is:

$$\frac{-\pi}{2}+\frac{\epsilon}{\sin\epsilon}+\ln(\cot(\epsilon/2))\tag{1}$$

And as $\epsilon\to 0^+$ (1) goes to $\infty.$ So the integral diverges.

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This is not surprising, because $\frac{\arcsin x}{x^2}$ is close to $\frac{1}{x}$ when $x$ is near zero, and $\int_{0}^{1}\frac{dx}{x}$ diverges.

This can be made rigorous by noting that $$f(x)=\frac{\arcsin x}{x^2}-\frac{1}{x}=\frac{\arcsin x-x}{x^2}$$ can be made continuous at $0$ by defining $f(0)=0.$ Then $$\int_0^{1} \frac{\arcsin x}{x^2}\,dx$$ converges if and only if $$\int_0^{1} \left(\frac{\arcsin x}{x^2}-f(x)\right)\,dx=\int_0^1 \frac{1}{x}\,dx$$ converges.

Answer 2

Hint: for $x\in(0,1)$ $$\arcsin(x) > x$$

Answer 3

Recall that as $x \to 0$ we have that $\arcsin (x) \sim x$ and therefore

$$\frac{\arcsin (x)}{x^{2}}\sim \frac 1x$$

therefore the integral diverges by limit comparison test with $\int_0^1 \frac 1x dx$.