A simple general expression for a definite integral

Question

Let $$I_k = \int_0^{\frac{\pi}{2}} \sin x \cos^{2k} x \sqrt{1 + \sin x} \, dx,$$ where $k = 0,1,2,\ldots$. I wish to find a simple general expression for $I_k$ in terms of $k$. Simple here is the operative word.

Making a tangent half-angle substitution of $t = \tan \frac{x}{2}$ produces $$I_k = 4\int_0^1 \frac{t(1 + t)^{2k + 1} (1 - t)^{2k}}{(1 + t^2)^{2k + \frac{5}{2}}} \, dt.$$

Trying to guess its general form one notices: \begin{align*} k = 0 : \quad I_0 &= 4\int_0^1 \frac{t(1 + t)}{(1 + t^2)^{\frac{5}{2}}} \, dt = 4 \left [\frac{(t - 1)(t^2 + t + 1)}{3(t^2 + 1)^{\frac{3}{2}}} \right ]_0^1 = \frac{4}{3}\\ k = 1 : \quad I_1 &= 4\int_0^1 \frac{t(1 + t)^3(1 - t)^2}{(1 + t^2)^{\frac{9}{2}}} \, dt = 4 \left [\frac{(t - 1)^3(11t^4 + 33t^3 + 52t^2 + 33t + 11)}{105(t^2 + 1)^{\frac{7}{2}}} \right ]_0^1 = \frac{44}{105}\\ k = 2 : \quad I_2 &= 4 \int_0^1 \frac{t(1 + t)^5(1 - t)^4}{(1 + t^2)^{\frac{13}{2}}} \, dt = 4\left [\frac{(1 - t)^5(211t^6 + 1055t^5 + 2593 t^4 + 3370t^2 + 1055t + 211)}{3465(t^2 + 1)^{\frac{11}{2}}} \right ]_0^1 = \frac{844}{3465} \end{align*} So it appears $$I_k = 4 \left [\frac{(t - 1)^{2k + 1}p_k(t)}{a_k(t^2 + 1)^{2k + \frac{3}{2}}} \right ]_0^1,$$ where $a_k$ is a positive integer and $p_k(t)$ is a symmetric polynomial of degree $(2k + 2)$. But what is $a_k$ and $p_k(t)$?

Alternatively, a closed-form expression in terms of the hypergeometric function can be found. It is $$I_k = \frac{2}{3\sqrt{2}} {}_2 F_1 \left (\frac{3}{2},-2k;\frac{5}{2};1 \right ) + \frac{1}{2^{2k + \frac{3}{2}}(2k+1)} {}_2F_1 \left (2k + 1,2k + \frac{5}{2};2k + 2; \frac{1}{2} \right ),$$ but this is hardly simple since ultimately $I_k$ is just a positive rational number. Perhaps this expression can be simplified in some way.

So my question is, can a simple expression for $I_k$ be found?

Answer 1

Too long for a comment.

I did some back-of-the-envelope computation and obtained the following conjecture:

$$ I_k = 2 \sum_{n=0}^{k+1} \sum_{l=n}^{\left\lfloor\frac{n+k+1}{2}\right\rfloor} \frac{(-1)^{n}}{2n+2k+1} \binom{n+k+1}{2l}\binom{l}{n}. $$

The idea is that the following identity is likely to hold for $|z| < 1$:

$$ \sum_{k=0}^{\infty} I_k z^{2k} = \frac{\alpha\operatorname{arctanh}(\alpha) + \beta \operatorname{arctanh}(\beta)}{z^2}, \quad \text{where} \quad \alpha, \beta = \sqrt{z\bigl(z \pm i\sqrt{1-z^2}\bigr)} $$

(I did not compute this result by myself, rather jammed it to my Mathematica and simplified the result.)

Expanding the right-hand side and reading out the coefficient of $z^{2k}$ gives the above formula. Here is a Mathematica code for generating the above sequence:

Table[2 Sum[(-1)^n/(2 n + 2 k + 1) Binomial[n + k + 1, 2 l] Binomial[l, n], {n, 0, k + 1}, {l, n, (n + k + 1)/2}], {k, 0, 5}]

The following compares the various representations for $I_k$'s:

Answer 2

I think that we could simplify the problem letting $$\sqrt{1+\sin(x)}=t \implies x=-\sin ^{-1}\left(1-t^2\right)\implies dx=\frac{2 t}{\sqrt{1-\left(1-t^2\right)^2}}\,dt$$ This makes $$I_k=2\int_1^{\sqrt 2}\left(t^2-1\right) t^{2 k+1} \left(2-t^2\right)^{k-\frac{1}{2}}\,dt$$ $$\color{blue}{I_k=\frac{\sqrt{\pi }\, \Gamma (2 k+1)-4^k \Gamma \left(2 k+\frac{3}{2}\right) \left((2 k-1) B_{\frac{1}{2}}\left(k+1,k-\frac{1}{2}\right)-4 k B_{\frac{1}{2}}\left(k+1,k+\frac{1}{2}\right)\right)}{\sqrt{2}\, \Gamma \left(2 k+\frac{5}{2}\right)}}$$ which is valid $\forall k\geq 0$.

For integer $k$, this generates the sequence $$\left\{\frac{4}{3},\frac{44}{105},\frac{844}{3465},\frac{7676}{45045},\frac{271372} {2078505},\frac{785244}{7436429},\frac{40395868}{456326325},\frac{343225244}{450 8102925},\frac{3309796628}{49589132175},\cdots\right\}$$

These sequences are not found in $OEIS$.