Evaluating $\lim_{r \to \infty} r^c \frac{\int_0^{\pi/2}x^r \sin x dx}{\int_0^{\pi/2}x^r \cos x dx}$

Question

The question is basically to find out the following limit($c$ is any real number) $$\lim_{r \to \infty} r^c \frac{\int_0^{\pi/2} x^r \sin x \, dx}{\int_0^{\pi/2}x^r \cos x \, dx}$$

I tried to use the property of definite integral and rewrote it as $$\lim_{r \to \infty} r^c \frac{\int_0^{\pi/2}(\pi/2-x)^r \cos x dx}{\int_0^{\pi/2}x^r \cos x dx}$$ and then expanded the numerator using binomial theorem.But this didnot take me to the answer.I also tried using integration by parts in the numerator but that too didnot helped me.Any hint to go ahead will be highly appreciated.Thanks.

Answer 1

In fact, integration by parts works in this case. For convenience, we write

$$ S(r) := \int_{0}^{\pi/2} x^r \sin x \, dx, \qquad C(r) := \int_{0}^{\pi/2} x^r \cos x \, dx. $$

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Idea. The intuition is that, since $x^r$ is largest at $x = \frac{\pi}{2}$, the behavior of integrand near this point will determine the growth speed of $S(r)$ and $C(r)$. Thus loosely speaking

$$ S(r) \approx \int_{0}^{\pi/2} x^r \, dx = \frac{(\pi/2)^{r+1}}{r+1}$$

and

$$ C(r) \approx \int_{0}^{\pi/2} x^r \left(\frac{\pi}{2} - x\right) \, dx = \frac{(\pi/2)^{r+2}}{(r+1)(r+2)}. $$

This forces that $c = -1$.

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Solution. We make the following two observations:

• Applying integration by parts, for $r > -1$ we have $$ C(r) = \frac{1}{r+1}S(r+1), \qquad S(r) = \frac{(\pi/2)^{r+1}}{r+1} - \frac{1}{r+1}C(r+1). \tag{*} $$

• Using the fact that sine and cosine are bounded by $1$, we have the following crude bound $$ |S(r)| \leq \frac{(\pi/2)^{r+1}}{r+1}, \qquad |C(r)| \leq \frac{(\pi/2)^{r+1}}{r+1}. $$

Thus if $r \gg 1$, then applying $\text{(*)}$ twice,

$$\frac{S(r)}{C(r)} = \frac{(\pi/2)^{r+1} - C(r+1)}{S(r+1)} = (r+2) \frac{(\pi/2)^{r+1} - C(r+1)}{(\pi/2)^{r+2}- C(r+2)}.$$

Dividing both the numerator and the denominator by $(\pi/2)^{r+1}$ and utilizing our crude bound,

$$\frac{S(r)}{C(r)} = (r+2) \frac{1 + \mathcal{O}(r^{-1})}{(\pi/2) + \mathcal{O}(r^{-1})}. $$

Therefore

$$ \lim_{r\to\infty} r^{-1} \frac{S(r)}{C(r)} = \frac{2}{\pi} $$

and $c = -1$ is only the possible choice.

Answer 2

Show that $$\int_0^{\pi/2}x^r \sin(x)dx=(\frac{\pi}{2})^{r+1}\int_0^1t^r \sin(\pi t/2)dt=(\frac{\pi}{2})^{r+1}a_r$$

And integrating by parts $$a_r=\frac{1}{r+1}-\frac{\pi}{2(r+1)}\int_0^1t^{r+1}\cos(\pi t/2)dt$$

It is now easy to see that $a_r$ is equivalent to $1/(r+1)$ as $r\to +\infty$.

In the same way:

$$\int_0^{\pi/2}x^r \cos(x)dx=(\frac{\pi}{2})^{r+1}\int_0^1t^r \cos(\pi t/2)dt=(\frac{\pi}{2})^{r+1}b_r$$

and we have $b_r=\frac{\pi}{2(r+1)}a_{r+1}$. It is easy to finish.