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I'm trying to prove the convergence of

$$\int^{\infty}_{0} \frac{\sin^2x}{x^a}dx$$

First we split the integral

$$\int^{1}_{0} \frac{\sin^2x}{x^a}dx+\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx$$

Since

$$\frac{\sin^2x}{x^a}\leq\frac{1}{x^a}$$

by the p-test, the term $\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx$ converges if and only if $a>1$.

And for the other term we use

$$\frac{\sin^2x}{x^a}\cong\frac{x^2}{x^a}=\frac{1}{x^{a-2}}$$

So that the integral converges if and only if $a-2<1 \to a<3$.

Summing up, $\int^{\infty}_{0} \frac{\sin^2x}{x^a}dx$ converges if $a \in (1,3)$, or else diverges.

Is my reasoning correct?

Thanks in advance for any suggestion!

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Assuming that $a>0$, your considerations for the integral on $[0,1]$ are fine: as $\sin^2(x)/x^a$ is non negative you can use the asymptotic comparison test and the integral is convergent on $(0,1]$ iff $a<3$.

On the other hand, I think that the inequality $$\frac{\sin^2x}{x^a}\leq\frac{1}{x^a}$$ tells you only that the integral on $[1,+\infty)$ is convergent for $a>1$ but does not help to say that it is not convergent for $0<a\leq 1$ ($1/x^2\leq 1/x$ for $x\geq 1$ does not imply that $\int_1^{\infty}1/x^2\ dx=+\infty$!).

Alternatively for that part use $\sin(x)^2=(1-\cos(2x))/2$. Then your integral becomes $$\int^{\infty}_{1} \frac{\sin^2x}{x^a}dx=\frac{1}{2}\int^{\infty}_{1} \frac{1}{x^a}dx-\frac{1}{2}\int^{\infty}_{1} \frac{\cos(2x)}{x^a}dx.$$ On the R.H.S., the first integral is convergent iff $a>1$ and the second integral is convergent iff $a>0$ because $$\int_{1}^{+\infty} \frac{\cos (2x)}{x^a} dx=\left[\frac{\sin (2x)}{ 2x^a}\right]_1^{+\infty}+ \frac{a}{2}\int_1^{+\infty} \frac{\sin( 2x)}{x^{a+1}} dx=\frac{\sin (2)}{ 2}+ \frac{a}{2}\int_1^{+\infty} \frac{\sin( 2x)}{x^{a+1}} dx$$ and $$\int_1^{+\infty} \frac{|\sin( 2x)|}{x^{a+1}} dx\leq \int_1^{+\infty} \frac{1}{x^{a+1}} dx<\infty.$$ Hence your integral is convergent on $[1,+\infty)$ iff $a>1$.

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If you attempt to use the comparison test with $\sin^2(x)/x^a$ and $x^2/x^a$, then for $a\in(-\infty,3]$ the integral does not converge in the latter function and therefore says nothing about the convergence of the actual integral. Instead, you could use the same approximation as the first integral and get that it converges for $a\in[1,\infty)$

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