Calculate the following integral $\int_0^{\pi/2} \frac{\sin^m x\,\mathrm{d}x}{\sin x + \cos x}$, $m=2k-1$

Question

At the moment I am studing the following integral

$$K(m,n)= \int_0^{\pi/2} \frac{\sin^m x\,\mathrm{d}x}{\sin^nx + \cos^nx}.$$

For integers $m$,$n$. The question regarding both $K(1,1)$ and $K(n,n)$ has been asked countless of times before on the site. With the usual solution of $x \mapsto \pi/2 - x$. I have tried to study the case $K(2k-1,1)$ for integers $k$. It seems this integral does not take any elementary form, but can it be expressed in a easier fashion?

Beneath is my work that proved more or less useless.. Assume that $m$ can be written on the form $2k-1$ where $k$ is some positive integer. Then we have from the binomial expansion that $$ \frac{a^n + b^n}{a+b} = a^{n-1} - a^{n-2}b + a^{n-3}b^2 + \cdots - a b^{n-2} + b^{n-1} $$ for odd $n$. By using the usual substitution one obtains that $$ K(m,1) = \frac{1}{2} \int_0^{\pi/2} \frac{\sin^mx + \cos^mx}{\cos x + \sin x} \mathrm{d}x = \frac{1}{2} \int_0^{\pi/2} \left( \sum_{k=1}^m \cos^{m-k}(x)\sin^{k-1}(x)\right)\mathrm{d}x $$ Since the integral is convergent for any positive $k$, both $\sin x$ and $\cos x$ live in $L^2$ we can switch the limits. By using the standard definition of the beta integral, the integral turns into $$ K(m,1) = \frac{1}{4} \sum_{k=1}^m \operatorname{B}(x,y) $$ With $2x-1=m-k$ and $2y-1=k-1$. But this does not seem any easier.. By using the symmetri $\operatorname{B}(y,x)=\operatorname{B}(x,y)$, and inputing $m = 2p-1$, the integral can also be written as $$ K(2p-1,1) = \frac{1}{4}\operatorname{B}\left(p^*,p^*\right) + \frac{1}{2}\sum_{k=1}^{p-1} \operatorname{B}(p-k^*,k^*) $$ With $p^*=p/2$ and $k^*=k/2$ to make the notation slightly better.

Are there any other perhaps more elementary way to find an expression for the integral? Is it possible to simplify this expression more?

Answer 1

This is not an answer but too long for a comment. Preliminary investigations reveal the following: $$K(m,1) = \begin{cases} a_m \pi -b_m & \text{ if }m \text{ is odd}\\ c_m \sqrt2 \tanh^{-1}\left(\dfrac1{\sqrt2}\right) + d_m & \text{ if }m \text{ is even} \end{cases}$$ where $a_m,b_m,c_m,d_m$ are non-negative rational numbers. We have $$(a_1,b_1) = (1/4, 0); (a_3,b_3) = (1/4, 1/4); (a_5,b_5) = (7/{32}, 1/4); (a_7,b_7) = (3/{16}, 5/{24}); (a_9,b_9) = (83/512, 1/6)$$

$$(c_0,d_0) = (1, 0); (c_2,d_2) = (1/2,0); (c_4,d_4) = (1/4, 1/6); (c_6,d_6) = (1/{8}, 1/{4});$$

Answer 2

Mathematica returns:

$$ \frac{2^{-m/2} \left(\sqrt{\pi } 2^{m/2} (m-1) \, _2F_1\left(\frac{1}{2},1;1-\frac{m}{2};-1\right) \Gamma \left(\frac{m}{2}\right)-\csc (\pi m) \Gamma \left(\frac{m+1}{2}\right) \left(2^{\frac{m}{2}+1} \, _2F_1\left(1,1;\frac{3-m}{2};-1\right) \sin (\pi m)+\sqrt{2} \pi (m-1)\right)\right)}{(m-1) \Gamma \left(\frac{m+1}{2}\right)} $$ I am not quite sure what to make of this, since $\csc(\pi m)$ is infinite. Now, partly it simplifies against the $\sin$ term, but partly it does not.