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Please help me in solving integral

$\int_0^1 y^{(1 + b)} \left(\frac{c + y}{1 + c y}\right)^b dy$, where $(0, $(0

I want the final solution in a simplified form not in terms of hypergeometric forms like Appell $F_1$ and $_2F_1$. Approximate solution. will also work.

Thanks

2 Answers 2

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Mathematica gives

$\frac{c^b \text{AppellF1}\left[2+b,-b,b,3+b,-\frac{1}{c},-c\right]}{2+b}$

I don't think it has a simpler form other than that. To do an approximation, maybe you need to specify some conditions on $b$ and $c$, such as $b \approx 0$ or $c \approx 0$.

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The interpolating polynomial gives a good approximation taking $x=0$,$x=1/2$ and $x=1$.

$f(x)={{{c}}^b}\left( {x - 1} \right)\left( {2x - 1} \right) - 4x\left( {x - 1} \right){\left( {\frac{{2c + 1}}{{c + 2}}} \right)^b} + x\left( {2x - 1} \right)$

so you can evaluate

$\int_0^1x^{1+b}f(x)dx$

and get an approximation. For $b$ fixed large and $c$ any value, the approximation is very good. Similarily for the other symmetric situation. For $c=0$ and $b<0.1$ the approximation is very bad, but for $c\neq 0$ we can push the values to $.9$ and things look good, for $c<.2;.3$ we get a not too good approximation.

Here you can see some images. $p$ is blue, the original function in red.

Bad $c=0$ case. $b$ ranges $0$ to $1$ in tenths of unity. enter image description here Good $b=1$ case. $c$ ranges $0$ to $1$ in tenths of unity. enter image description here Not that bad $c=0.1$ case. $b$ ranges $0$ to $1$ in tenths of unity. enter image description here Good $c=0.6$ case. $c$ ranges $0$ to $1$ in tenths of unity.

enter image description here Good $b=0.62$ case. $c$ ranges $0$ to $1$ in tenths of unity.

enter image description here