Evaluate the given integral $$\int (1-x^{2008})^{\frac{1}{2007}} (1-x^{2007})^{\frac{1}{2008}} dx$$

Using integration by parts is out of equation because we can't integrate any of two brackets. I also cannot think of any substitution that can lead to integration. Could someone help me with this?

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    $\begingroup$ Very high exponents. Looks like a contest problem - is it, and from where if so? $\endgroup$ – Parcly Taxel May 13 '17 at 4:44
  • $\begingroup$ $x^2,(1-x)^2$ so on. $\endgroup$ – Takahiro Waki May 13 '17 at 6:03
  • $\begingroup$ This integral is only defined for $-1 \leq x \leq 1$. $\endgroup$ – Zaid Alyafeai May 13 '17 at 15:51
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    $\begingroup$ Seems that it is an integral of the form $\int f(x)f^{-1}(x)\;dx$. Any known trick for that? $\endgroup$ – Benedict W. J. Irwin Oct 17 '17 at 13:02
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    $\begingroup$ If the integral would have been from $0$ to $1$, and there would be addition in front of the factors, then the result would be $2$. I agree with the question of @ParclyTaxel, where is this from? $\endgroup$ – mickep Nov 11 '17 at 19:03

I will not give step-by-step calculations of integrals here since they are rather long and tedious.

As @BenedictWilliamJohnIrwin suggested, the integral is of the form $$\int f(x)f^{-1}(y)\, dx \tag1$$ where $y=f(x)=(1-x^{2008})^\frac1{2007}$.

Now it is known that $$\int f^{-1}(y)\, dy=xf(x)-\int f(x) \, dx \tag2$$ (see http://www.med.umich.edu/schnell-lab/assets/sm6.pdf for details) so on integrating $(1)$ by parts and combining it with $(2)$, we get $$\left[xf(x)^2-f(x)\int f(x)\, dx\right]-\int \left(xf(x)f'(x)-\int f(x)\, dx\right) dx \tag3$$ We want to write it in this form since we do not need to deal with $(1-x^{2007})^\frac1{2008}$.

The remainder of the calculations relies heavily on the use of WolframAlpha. The final result is very messy since it contains many hypergeometric functions, so I will not display it.

By WolframAlpha, $$\int f(x) \, dx=\frac{x}{4015}\left(2008\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)+2007(1-x^{2008})^\frac1{2007}\right)+C_1 \tag4$$ where $C_1$ is a constant.

$(3)$ can be written as $$g_1(x)-\int \left(g_2(x)-\int f(x) \, dx\right)dx$$ where $g_1(x)$ is $$\left[x(1-x^{2008})^\frac2{2007}-\frac{1}{4015}x(1-x^{2008})^\frac1{2007}\left(2008\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)+2007(1-x^{2008})^\frac1{2007}\right)\right]=x(1-x^{2008})^\frac1{2007}\left[\frac{2008}{4015}\left((1-x^{2008})^\frac1{2007}-\, _2F_1\left(\frac1{2008}, \frac{2006}{2007}; \frac{2009}{2008}; x^{2008}\right)\right)\right]$$ and $g_2(x)$ is $$x(1-x^{2008})^\frac1{2007}\cdot\left(-\frac{1}{2007}\right)(1-x^{2008})^{-\frac{2006}{2007}}\cdot 2008x^{2007}=-\frac{2008}{2007}x^{2008}(1-x^{2008})^{-\frac{2005}{2007}}.$$

Again using WolframAlpha we find that $\int g_2(x)\, dx$ equates to $$\frac{2008}{6023}x\left((1-x^{2008})^\frac2{2007}-\, _2F_1\left(\frac1{2008}, \frac{2005}{2007}; \frac{2009}{2008}; x^{2008}\right)\right)+C_2$$ where $C_2$ is a constant.

Finally, we integrate $(4)$. To do this, we split the integrand into two functions; that is, $\int f(x)\, dx = g_3(x)+ g_4(x)+C_1$ where $g_4(x)=\frac{2007}{4015}xf(x)$ and $g_3(x)$ is the term with the hypergeometric function.

Using the second entry of http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/21/01/02/01/ with $\alpha = \frac{2009}{2008}$ (and WolframAlpha), the integral of $g_3(x)$ is $$-\frac{2008^2}{2009(4015)}x^{\frac{2009}{2008}}\, _3F_2\left(\frac1{2008}, \frac{2006}{2007}, \frac{2009}{2008}; \frac{2009}{2008}, \frac{4017}{2008}; x^{2008}\right)+C_3$$ and on integrating by parts (using $(4)$), the integral of $g_4(x)$ is $$-\frac{2007}{4015(6022)}x^2\left(2007(1-x^{2008})^\frac1{2007}+1004\, _2F_1\left(\frac1{1004}, \frac{2006}{2007}; \frac{1005}{1004}; x^{2008}\right)\right)+C_4$$ wherre $C_3$ and $C_4$ are constants.

Combining the relevant results gives the integral and some constant $C$. Don't forget to integrate $C_1$!

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    $\begingroup$ Note to $(2)$. $\enspace$ NO: $\int f^{-1}(x)\, dx=xf(x)-\int f(x) \, dx \enspace$ YES: $\int f^{-1}(y)\, dy |_{y=f(x)} = xf(x)-\int f(x) \, dx $ $\endgroup$ – user90369 Nov 13 '17 at 9:28
  • $\begingroup$ @user90369 thanks - have edited. $\endgroup$ – TheSimpliFire Nov 13 '17 at 19:07
  • $\begingroup$ Note to $(1)$ : BenedictWilliamJohnIrwin has written something else, please compare. $\endgroup$ – user90369 Nov 14 '17 at 7:25

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