$\int_{0}^{1}\frac{\sin(x^p)}{x}dx$ values of convergence

Question

Consider the integral $\int_{0}^{1}\frac{\sin(x^p)}{x}dx$ for $p>0$. Then for which values of $p$, does the integral converge?

So we know that for $x=0$, the integral is undefined. So I should make it $\lim_{t \to 0^+}$. After that, I evaluated the integral $\lim_{t \to 0^+}\int_{t}^{1}\frac{\sin(x^p)}{x}$. My answer is that $p>0$ is the condition for the integral to converge. Is this right?

Answer 1

The integral is not undefined for $x=0$. Use the series expansion of $\sin(t)$, replace $t$ by $x^p$ to get $$\frac{\sin(x^p)}x=x^{p-1}-\frac{1}{6} x^{3 p-1}+\frac{1}{120} x^{5 p-1}+\cdots$$ $$\int_0^1 \frac{\sin(x^p)}x dx\sim\frac{1703}{1800 \,p}\qquad \text{if} \qquad p >0$$

Answer 2

Consider the change of variable $$u=x^p,\qquad x=u^{1/p}$$ then $dx=\frac{1}{p} u^{(1/p)-1}\,du$. It follows that

1. For $p>0$ $$ I_p=\int^1_0\frac{\sin x^p}{x}\,dx =\frac{1}{p}\int^1_0\frac{\sin u}{u^{1/p}}\frac{1}{u^{1-(1/p)}}=\frac{1}{p}\int^1_0\frac{\sin u}{u}\,du $$ In this case $I_p$ integral exists for all $p>0$ (the function $u\mapsto \frac{\sin u}{u}$ for $u\neq0$ and $1$ for $u=0$, being continuous, is integrable in $[0,1]$).
2. For $p=0$ then $I_p=\int^1_0\frac{1}{x}\,dx=\infty$.
3. If $p<0$ then $$ I_p=-\frac{1}{p}\int^\infty_1\frac{\sin x}{x}\,dx $$ In this case $I_p$ converges as an improper integral, that is $$-\frac{1}{p}\lim_{a\rightarrow\infty}\int^a_1\frac{\sin x}{x}\,dx$$ exists and is finite. However, $\frac{\sin x}{x}$ is not integrable over $[1,\infty)$ in the sense of Lebesgue.

Answer 3

If $p>1$, then you can use L'Hopital's rule to see that $\frac{\sin(x^p)}{x}$ is continuous on $[0,1]$ and is therefore bounded. Thus the integral converges if $p>1$.

(It would be more accurate to stay it extends to a continuous function on $[0,1]$ since it's technically not defined at $x=0$.).

If $0 < p < 1$, then

$$\int_0^1 \frac{\sin(x^p)}{x} \ dx = \int_0^1 \frac{\sin(x^p)}{x^p x^{1-p}}\ dx = \int_0^1 \frac{x^{p-1}\sin(x^p)}{x^p}\ dx $$

and you can use $u$-substitition with $u = x^p$ to show that this also converges.

Answer 4

If $p>0$, then $$ 0 < \int_0^1 {\frac{{\sin (x^p )}}{x}dx} = \int_0^1 {x^{p - 1} \frac{{\sin (x^p )}}{{x^p }}dx} \leq \int_0^1 {x^{p - 1} dx} = \frac{1}{p} < + \infty , $$ since $\frac{\sin w}{w}\leq 1$ for all $w \geq 0$.