I would like to compute the integral:
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$$
where $$ a<b$$ and $$ p\in\mathbb{N}$$
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx=\frac{2}{b-a} \int_a^b \frac{x^p \mathrm dx}{\sqrt{1-\left(\frac{2x-(a+b)}{b-a}\right)^2}} $$
The substitution $x\rightarrow \frac{2x-(a+b)}{b-a}$ gives:
$$ \frac{1}{2^p} \int_{-1}^{1} \frac{((b-a)x+a+b)^p}{\sqrt{1-x^2}} \mathrm dx$$
I tried to make an integration by parts:
$$ \frac{1}{2^p}\left(2^{p-1}\pi(a^p+b^p)-p(b-a) \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx \right)$$
What about the integral $$ \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx $$?
Best Answer
Let $x = a + (b-a)t$. Then $$I(p) = \int_0^1 \dfrac{(a+(b-a)t)^p}{\sqrt{t(1-t)}} dt$$ We have that $$\int_0^1 \dfrac{t^k dt}{\sqrt{t(1-t)}} = \sqrt{\pi} \dfrac{\Gamma(k+1/2)}{\Gamma(k+1)}$$ The above simplification is possible due to the following reason. Recall that the $\beta$ function is defined as $$\beta(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$$ The $\beta$-function is closely related to the $\Gamma$ function through the relation $$\beta(x,y) = \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$$ The proof for the above claim can be seen here.
Hence, we get that $$\int_0^1 \dfrac{t^k dt}{\sqrt{t(1-t)}} = \int_0^1 t^{k-1/2} (1-t)^{-1/2} dt = \int_0^1 t^{(k+1/2)-1} (1-t)^{1/2-1} dt\\ = \beta(k+1/2,1/2) = \dfrac{\Gamma(k+1/2) \Gamma(1/2)}{\Gamma(k+1)} = \sqrt{\pi} \dfrac{\Gamma(k+1/2)}{\Gamma(k+1)}$$ Further, $$\Gamma(1/2) = \sqrt{\pi}$$ You can look up here for some particular values of the $\Gamma$ function. $$\Gamma(k+1/2) = \dfrac{(2k-1)(2k-3)\cdots 1}{2^k} \sqrt{\pi} \text{ where } k \in \mathbb{Z}^+$$ The above is so since $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(1/2) = \sqrt{\pi}$. $$\Gamma(k+1) = k! \text{ where } k \in \mathbb{Z}^+$$ $$\dfrac{\Gamma(k+1/2)}{\Gamma(k+1)} = \dfrac{(2k-1)(2k-3)\cdots 1}{2^k k!} \sqrt{\pi} = \dfrac1{4^k} \binom{2k}{k} \sqrt{\pi}$$
Hence, $$I(p) = \int_0^1 \sum_{k=0}^{p} \binom{p}{k} a^{p-k}(b-a)^{k} \dfrac{t^{k}}{\sqrt{t(1-t)}} dt\\ = \displaystyle \sum_{k=0}^{p} \left(\binom{p}{k} a^{p-k}(b-a)^{k} \int_0^1 \dfrac{t^{k}}{\sqrt{t(1-t)}} dt \right)\\ = \pi \times \left( \sum_{k=0}^{p} \left( \binom{p}{k}\binom{2k}{k} \dfrac{a^{p-k}(b-a)^{k}}{4^k} \right) \right)$$
We have that $$I(0) = \pi$$ $$I(1) = \dfrac{a+b}{2} \pi$$ $$I(2) = \dfrac{3a^2+2ab+3b^2}{8} \pi$$ $$I(3) = \dfrac{5a^3+3a^2b+3ab^2+5b^3}{16} \pi$$
I am not sure if you can simplify this further in terms of elementary functions.