What Real Number and Positive Limit Solve This Integral Ratio Challenge?

[Ackbach]

Here is this week's POTW:

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Find a real number $c$ and a positive number $L$ for which
$$\displaystyle\lim_{r\to\infty} \frac{\displaystyle r^c \int_0^{\pi/2} x^r \sin(x) \,dx}{\displaystyle \int_0^{\pi/2} x^r \cos(x) \,dx} = L.$$

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[Ackbach]

Congratulations to Kiwi for his correct solution, which follows:

Spoiler

I choose to let r tend to infinity through the sequence of integer values r = {3,7,11, ..., 3+4t ...}.

Let
\(I= \int x^r \sin(x) \,dx\)

and
\(J = \int x^r \cos(x) \,dx\)

so
\(\displaystyle\lim_{r\to\infty} \frac{ r^c \int_0^{\pi/2} x^r \sin(x) \,dx}{ \int_0^{\pi/2} x^r \cos(x) \,dx} = \displaystyle\lim_{r\to\infty} \frac{r^c ^{\pi/2}_0}{[J]^{\pi/2}_0}L.\)

Now by repeated application of integration by parts (remembering r = 3 mod 4):
\(J=x^r \sin(x) + r x^{r-1} \cos(x) - r^2 x^{r-2} \sin(x) - r^3 x^{r-3} \cos(x) +
... + r^{r-3} x^3 \sin(x) + r^{r-2} x^2 \cos(x) - r^{r-1} x \sin(x) - r^r \cos(x)\)

\([J]^{\pi/2}_{0}=
(x^r \sin(x) - r^2 x^{r-2} \sin(x) + ... + r^{r-3} x^3 \sin(x) - r^{r-1} x \sin(x))|_{x=\pi/2} - \displaystyle\lim_{x\to 0} \frac rx
(x^r \cos(x) - r^2 x^{r-2} \cos(x) + ... + r^{r-3} x^3 \cos(x) - r^{r-1} x \cos(x))\)

\(\therefore [J]^{\pi/2}_{0}=
(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x )|_{x=\pi/2} - \displaystyle\lim_{x\to 0} \frac rx
(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x )\)

\(\therefore [J]^{\pi/2}_{0}=
(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x )|_{x=\pi/2} - 1\) \(\tag{1}\)

Similarly
\(I=-x^r \cos(x) + r x^{r-1} \sin(x) + r^2 x^{r-2} \cos(x) - r^3 x^{r-3} \sin(x) + ...
-r^{r-3}x^3 \cos(x) + r^{r-2} x^2 \sin(x) + r^{r-1} x \cos(x) - r^r \sin(x)\)

noting that the cos terms are all zero for x=0 or x=pi/2 and that the sin terms are not singular and are equal to zero as x approaches 0.
\(^{\pi/2}_{0}= \frac rx
(x^r \sin(x) - r^2 x^{r-2} \sin(x) + ... + r^{r-3} x^3 \sin(x) - r^{r-1} x \sin(x))|_{x=\pi/2}\)

\(\therefore ^{\pi/2}_{0}= \frac rx
(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x )|_{x=\pi/2}\) \(\tag{2}\)

So finally, using (1) and (2):
\(\displaystyle\lim_{r \to \infty} \frac{r^c^{\pi/2}_0}
{ [J]^{\pi/2}_0}=\displaystyle\lim_{r \to \infty}\frac {r^c(\frac rx(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x ))|_{x=\pi/2}}{(x^r - r^2 x^{r-2} + ... + r^{r-3} x^3 - r^{r-1} x )|_{x=\pi/2}-1}=\displaystyle\lim_{r \to \infty}\frac{r^cr2}{\pi}\)

So
\(\displaystyle\lim_{r\to\infty} \frac{ r^c \int_0^{\pi/2} x^r \sin(x) \,dx}{ \int_0^{\pi/2} x^r \cos(x) \,dx} =\frac 2\pi\) if c = -1 and cannot be a positive real number for any other value of c.