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The question I am working on: Evaluate

$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$

So using integration by parts to solve:

(letting $u=(1-x)^4$ and $dv=x^4$)

$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$

Is it correct so far? If so ...

$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$

$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$

But

$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$


Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around?

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    (Sort of) related: http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-1062012-04-09

2 Answers 2

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Using the binomial theorem and then long division, $ \frac{x^4(1-x)^4}{1+x^2}= \frac{x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2}= x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2} $ so that $ \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx= \frac17-\frac46+\frac55-\frac43+\frac41-4\arctan\frac\pi4 =\frac{22}7-\pi\approx\frac1{790.~833~125~927~563} $ There's also a problem with your integration by parts since $ \eqalign{ u &=& (1-x)^4=(x-1)^4 &\qquad& v &=& x^4 \\\\ du &=& 4(x-1)^3 \, dx &\qquad& dv &=& \frac15 x^5 dx } $ so $ \eqalign{ I_1 &= \int_0^1 x^4 (1-x)^4 dx \\ &= \int u\,dv = uv - \int v\,du \\ &= \left[ \frac45 x^5 (1-x)^4 \right]_0^1 + \frac45 \int_0^1 x^5 (1-x)^3 dx \\ &= \frac45 \int_0^1 x^5 (1-x)^3 dx \\ &= \frac45\cdot\frac36 \int_0^1 x^6 (1-x)^2 dx \\ &= \frac45\cdot\frac36\cdot\frac27 \int_0^1 x^7 (1-x) dx \\ &= \frac45\cdot\frac36\cdot\frac27\cdot\frac18 \int_0^1 x^8 dx \\ &= \frac45\cdot\frac36\cdot\frac27\cdot\frac18\cdot\frac19 = \frac1{630} } $ where we had to apply integration by parts repeatedly using $u=(1-x)^n,~dv=x^m\,dx$ $du=-n(1-x)^{n-1}\,dx,~v=\frac{x^{m+1}}{m+1}\implies$ $\int x^m(1-x)^ndx=\frac{x^n(1-x)^{m+1}}{m+1} +\frac{n}{m+1}\int x^{m+1}(1-x)^{n-1}dx$ until the powers of $(1-x)$ went away. In fact, we just made a special case of the calculation $ \int_0^1 x^a (1-x)^b = B(a+1,b+1) = \frac{a!~b!}{(a+b+1)!} $ of the well-known Beta function for $a=b=4$ (our method works for $a,b\in\mathbb{N}$ and, with an adaptation of the above formula using the Gamma function, also for real $a,b\ge0$).

Now we can see that the reciprocal of the central integral is certainly beween $630$ and $1260$.

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    @DavidMitra: Actually I had $du$ right but used it incorrectly, which is what you surely noticed. For the general step, $u=(1-x)^n,~dv=x^m\,dx$ $du=-n(1-x)^{n-1}\,dx,~v=\frac{x^{m+1}}{m+1}\implies$ $\int x^m(1-x)^ndx=\frac{x^n(1-x)^{m+1}}{m+1}+\frac{n}{m+1}\int x^{m+1}(1-x)^{n-1}dx$2012-04-09
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Hint :

Rewrite integral into form :

$I=\int \frac{x^8+x^6}{1+x^2} \,dx + \int \frac{x^6+x^4}{1+x^2} \,dx +4\cdot \int \frac{x^6}{1+x^2} \,dx -4\cdot \int \frac{x^7+x^5}{1+x^2} \,dx$

for third integral do long division .

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    @pedja: I'm married (QED), which implicitly constitutes not only a constructive proof of my mortality, but comes also with canonical induced lift and pair of projections which alas don't commute.2012-04-10