# finding integral

I have trouble with Calculation of the following integral: for $0<t<\sqrt 6$ and $p>1$

$$\int_0^{\sqrt 6} \left(\frac{t^4}{390}-\frac{t^2}{6}+1\right)^p\;dt.$$

Which substitution would work here?

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Nothing pleasant unfortunately. If $p$ is a positive integer we can expand and integrate term by term. If $p$ is not an integer, we don't even have that, and should use a numerical method. – André Nicolas Dec 22 '11 at 23:57
cross posting – The Chaz 2.0 Dec 23 '11 at 0:05
If $p$ is a half-integer, your integral is expressible in terms of the elliptic integrals. – J. M. Dec 23 '11 at 0:27
It quite depends on p. – Strin Dec 23 '11 at 9:27
Start simplifying it to $$\left(\left(\frac{t^2}{12}+1 \right)^2 - \left(\frac{t^2}{12}\frac{\sqrt{41}}{\sqrt{65}}\right)^2\right)^p$$ – Kirthi Raman Mar 3 '12 at 16:52

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_0^{\root{6}}\pars{{t^{4} \over 390} - {t^{2} \over 6} + 1}^{p}\,\dd t:\ {\large ?}.\quad p > 1.\quad}$ Note that $\ds{\root{6} \approx 2.4495}$.
Expression $\pars{1}$ is reduced to: \begin{align} &\int_0^{\root{6}}\pars{{t^{4} \over 390} - {t^{2} \over 6} + 1}^{p}\,\dd t =\int_{0}^{\root{6}}\pars{1 - {t^{2} \over \mu^{2}}}^{p} \pars{1 - {t^{2} \over \nu^{2}}}^{p}\,\dd t \\[3mm]&=\int_{0}^{6}\pars{1 - {t \over \mu^{2}}}^{p} \pars{1 - {t \over \nu^{2}}}^{p}\,\pars{\half\,t^{-1/2}\dd t} \\[3mm]&=\half\root{6}\int_{0}^{1}t^{-1/2}\pars{1 - {6 \over \mu^{2}}\,t}^{p} \pars{1 - {6 \over \nu^{2}}\,t}^{p}\,\dd t \end{align} The result is related to the Appell Hypergeometric Series. In order to show clearly that relation we rewrite the above result as: \begin{align} &\int_0^{\root{6}}\pars{{t^{4} \over 390} - {t^{2} \over 6} + 1}^{p}\,\dd t \\[3mm]&=\root{6}\times \\[3mm]&\bracks{% \overbrace{{\Gamma\pars{\color{#00f}{3/2}} \over \Gamma\pars{\color{#c00000}{1/2}} \Gamma\pars{\color{#00f}{3/2} - \color{#c00000}{1/2}}}}^{\ds{=\ \half}}\ \int_{0}^{1}t^{\color{#c00000}{1/2} - 1}\ \overbrace{\pars{1 - t}^{\color{#00f}{3/2} - \color{#c00000}{1/2} - 1}}^{\ds{=\ 1}} \pars{1 - {6 \over \mu^{2}}\,t}^{-\pars{-p}} \pars{1 - {6 \over \nu^{2}}\,t}^{-\pars{-p}}\!\!\!\!\!\dd t} \\[3mm]&=\root{6}\ {\rm F}_{1}\pars{% \color{#c00000}{\half},-p,-p,\color{#00f}{3 \over 2},{6 \over \mu^{2}}, {6 \over \nu^{2}}} \end{align} $\ds{{\rm F}_{1}}$ is an Appell Hypergeometric Series. $$\mbox{As}\quad {6 \over \mu^{2}} = {1 \over 130}\,\pars{65 + \root{2665}} \quad\mbox{and}\quad {6 \over \nu^{2}} = {12 \over 65 + \root{2665}}$$