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I am considering integrals of the form $$ I_{n,m}=\int_{-y_0}^{y_0}y^m\left(1-\frac{y^2}{y_0^2}\right)^{\frac{n}{2}}\text{d}y, $$ where $m\in \mathbb{N}_{>0}$ and $n=2k+1$, with $k\in\mathbb{N}$. Is there a simple substitution to solve these? At first I thought of $y=y_0\cos\theta$, but this gives me fractional powers of trigonometric functions, which I do not know how to calculate.

I am pretty sure there should be a general solution possible, as Mathematica gives in the case $m=2$, $n=3$ the result $I_{2,3} = y_0^3\pi/16$.

Funzies
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1 Answers1

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You can sub $y=y_0 u$ and get that the integral is

$$y_0^{m+1} \int_{-1}^1 du \, u^m \, (1-u^2)^{n/2}$$

Split the integral in two:

$$\begin{align}& y_0^{m+1} \int_{-1}^0 du \, u^m \, (1-u^2)^{n/2} + y_0^{m+1} \int_{0}^1 du \, u^m \, (1-u^2)^{n/2}\\ &=(-1)^m y_0^{m+1} \int_{0}^1 du \, u^m \, (1-u^2)^{n/2} + y_0^{m+1} \int_{0}^1 du \, u^m \, (1-u^2)^{n/2}\\ &= \frac12 [1+(-1)^m]\, y_0^{m+1} \underbrace{\int_0^1 dv \, v^{(m-1)/2} (1-v)^{n/2}}_{\text{Beta function}}\\&= \frac12 [1+(-1)^m]\, y_0^{m+1}\frac{\Gamma \left (\frac{m+1}{2} \right )\Gamma \left (\frac{n}{2}+1 \right )}{\Gamma \left (\frac{m+n+3}{2} \right )}\end{align}$$

The integral should be zero when $m$ is odd. This expression also agrees with your specific result.

Ron Gordon
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