When Does The Integral Converge? - Part 2

First of all, if we flip the integrand arround the $x = \frac{\pi}{2}$ line, we get

$\large \int_{0}^{\pi} \left( \sin (\pi - x) \right)^{k\cos (\pi - x)} dx = \int_{0}^{\pi} \left( \sin x \right)^{-k\cos x} dx$

So, if $k = p$ makes it converge, then also does $k = -p$ . We'll restrict ourselves to looking only at this flipped form of the integral, for positive k.

Also, it's not hard to show, that, in this case, for positive k, the integrand $\displaystyle \left( \sin x \right)^{-k\cos x}$ is bounded when $x \in (1,\pi)$ , so we'll only look at the integral from $0$ to $1$ .

To find the places where the integral converges, we'll make upper and lower bounds for the integrand, and to do that, we'll use these upper and lower bounds for $\sin$ and $\cos$ :

$\displaystyle \frac{x}{2} \leq \sin (x) \leq x$ Here, the lower bound is very rough, but it will do the job, when necessary.

$\displaystyle 1 - \frac{x^2}{2} \leq \cos (x) \leq 1 - \frac{x^2}{3}$

We can use these bounds to say:

$\large \int_{0}^{1} x ^{-k\left(1 - \frac{x^2}{2}\right)} dx < \int_{0}^{1} \left( \sin x \right)^{-k\cos x} dx < \int_{0}^{1} \left( \sin x \right)^{-k\left(1 - \frac{x^2}{3}\right)} dx$

And rearragne:

$\large \int_{0}^{1} \frac{x^{\frac{kx^2}{2}}}{x^k} dx < \int_{0}^{1} \left( \sin x \right)^{-k\cos x} dx < \int_{0}^{1} \frac{(\sin x)^{\frac{kx^2}{3}}}{(\sin x)^k} dx$

Now, its not hard to see that, in the integration interval $(0,1)$ , the following holds:

• $\displaystyle \large (\sin x)^{\frac{kx^2}{3}} \leq 1$

• There is some $\displaystyle \large p \in (0,1) | p \leq x^{\frac{kx^2}{2}} \leq 1$

This means we can extend the upper and lower bounds even further:

$\large \int_{0}^{1} \frac{p}{x^k} dx < \int_{0}^{1} \left( \sin x \right)^{-k\cos x} dx < \int_{0}^{1} \frac{1}{(\sin x)^k} dx$

Finally, let's use the lower bound for the $\sin$ :

$\large \int_{0}^{1} \frac{p}{x^k} dx < \int_{0}^{1} \left( \sin x \right)^{-k\cos x} dx < \int_{0}^{1} \frac{2^k}{x^k} dx$

$\large p \int_{0}^{1} \frac{1}{x^k} dx < \int_{0}^{1} \left( \sin x \right)^{-k\cos x} dx < 2^k \int_{0}^{1} \frac{1}{x^k} dx$

Therefore, we see that the integral converges if and only if the following integral converges:

$\large \int_{0}^{1} \frac{1}{x^k} dx$

But it's well known this integral converges, in the our case, for $k \in [0,1)$ . Thus, the original integral converges for $k \in [0,1)$ and also for $k \in (-1,0)$ , due to the lemma we used in the beginning.

We have, finally: $(a,b) = (-1,1)$ , and $b-a = 2$

Case 1 : We are trying to find an upper bound( $b$ ). For $k>0$ we can see that at $x=π$ the integrand goes to infinity. So, we can check whether the integral will converge or not by $\mu$ -test.

If there exists some $\mu\leq1$ such that $\lim_{x\to π-} (sin(x))^{kcos(x)}(π-x)^{\mu}$ exists, then the integral is convergent.

$\lim_{x\to {π-}} (sin(x))^{kcos(x)}(π-x)^{\mu}$ =

$\ lim_{x\to {π-}} (sin(π-x))^{kcos(x)}(π-x)^{\mu}$ =

$\lim_{x\to {π-}} (π-x)^{\mu -k}$ is finite ( $cos(π)=-1$ ).

As $\mu <1 \Rightarrow k\leq 1$ . Or, $b=1$ .

Case 2 : Similarly for $k<0$ we can show that $a=-1$

Hence, $b-a=2$ .