For what values of $p>0, \quad \int^{1}_{0} \frac{x}{\sin{(x^{p})}} \operatorname d\!x$ converges?

Question:

For which $p>0$ does the improper integral $$\displaystyle I = \int^{1}_{0} \frac{x}{\sin{(x^{p})}} \ dx$$ exist?

My thoughts:

Initially we'll let $\displaystyle I_{t} = \int^{1}_{t} \frac{x}{\sin{(x^{p})}} \ dx$ hence $\displaystyle I = \lim_{t\to 0} I_{t}$.

Now we must evaluate the integral for $p=1$ and $p\not=1$ but I have no idea how to evaluate either case.

• What do you know about the behavior of $\sin u$ for $|u|$ "small"? – Andrew D. Hwang Feb 16 '14 at 16:59
• "Now we must evaluate the integral" No, we must only determine whether it converges or not, which is not the same kind of fish. – Did Feb 16 '14 at 17:02
• There isn't any oscillation near $0$ to prevent the integral to blowing up as you make $t$ apporach $0$. So my guess is that this exists whenever $0< p <2$ and diverges for all $p \geq 2$. That this blows up for $p\geq 2$ can easily be proved using the estimate $\sin(x) \leq x$. That it converges for $0<p<2$ can be shown using the estimate $\sin(x) \geq 2x/\pi$ (for $0\leq x \leq \pi/2$). – i like xkcd Feb 16 '14 at 17:06
• So what would I do to work out $p$ then? – user2850514 Feb 16 '14 at 17:31
• I have never done anything like this before, what am I meant to do to find $p$? – user2850514 Feb 16 '14 at 17:52

For $0<x<\pi/2$ the function $f(x)=\sin x$ is increasing, with $0<f'(x)<1$. Consequently $0<f(x)< x$ in this range. Also, $f''(x)<0$ on this interval, which means the function is concave down. In particular, it lies above its secant line through $(0,0)$ and $(\pi/2,1)$. Concretely stated, $\sin x > \frac{2}{\pi}x$ in the interval $(0,\pi/2)$.

Armed with the above inequalities, you can state that $$\frac{1}{x^{p-1}}=\frac{x}{ x^p}<\frac{x}{\sin (x^p)}< \frac{\pi}{2}\frac{x}{ x^p} = \frac{\pi}{2}\frac{1}{x^{p-1}}$$ Since the integral $\displaystyle \int_0^1 \frac{1}{x^{p-1}}\,dx$ converges if and only if $p-1<1$, the same is true for the original integral (comparison test).

Alternatively, use Limit Comparison Test (this is more likely to be expected of students in a calculus course): Since $$\lim_{t\to 0} \frac{\sin t}{t} =1$$ it follows that $$\lim_{x\to 0}\left(\frac{x}{\sin (x^p)} \bigg/ \frac{1}{ x^p }\right)=1$$ which allows the test to be applied.

A common approach to improper integrals is to split off the singularity into a separate summand that is easy to integrate.

In this case, with the singularity at $x=0$, we have $\sin x \sim x$, and so $1/\sin(x^p) \sim 1/x^p$. Using this approach, we should rewrite the integral as

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

It's not to hard to show the second summand can be continuously extended to $0$, and so its integral would be proper, and thus

$$\cdots = \int_0^1 x^{p-1} \, dx + \int_0^1 \left( \frac{x^p - \sin(x^p)}{x^p \sin(x^p)} \right) x \, dx$$

and now it's easy to determine whether or not the integral converges!

Incidentally, this is also a nice trick when you're trying to numerically estimate an integral: the singularity of the original integrand (when $p>1$) makes it ill-suited for numerical estimation. After rewriting it, though, we can evaluate the singular part exactly, and the remaining integrand is much more well-behaved.

If you make a change of variable such that $x^p=y$, the integrand becomes $$\frac{y^{\frac{2}{p}-1} \csc (y)}{p}$$ Now, if $csc(y)$ is now developed as an infinite Taylor series built at $y=0$, it becomes $$\frac{y^{\frac{2}{p}-2}}{p}+\frac{y^{2/p}}{6 p}+\frac{7 y^{\frac{2}{p}+2}}{360 p}+\frac{31 y^{\frac{2}{p}+4}}{15120 p}+ ...$$ and the integral has finite values only if $0<\Re(p)<2$.