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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 ...

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

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


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

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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What do you mean by evaluating the inequality? It is certainly true since $1\le1+x^2\le2$ on $[0,1]$. Do you really need the central integral? If so, note that $$\frac{x^4(1-x)^4}{1+x^2}=x^4(x-1)^2\left(1-\frac{2x}{1+x^2}\right)$$ while you can use $\int\frac{2x}{1+x^2}=\ln(1+x^2)+c$ and $\int\frac{1}{1+x^2}=\arctan(x)+c$ to integrate the linear and constant remainder terms from the general polynomial quotient. – bgins Apr 9 '12 at 12:11
(Sort of) related:… – Hans Lundmark Apr 9 '12 at 14:21
up vote 6 down vote accepted

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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@TMM, thanks, I fixed it. – bgins Apr 9 '12 at 13:21
$du=-4(x-1)^3$. – David Mitra Apr 9 '12 at 13:23
@DavidMitra: Actually I had $du$ right but used it incorrectly, which is what you surely noticed. For the general step, $$u=(1-x)^n,~dv=x^m\,dx$$ $$du=-n(1-x)^{n-1}\,dx,~v=\frac{x^{m+1}}{m+1}\implies$$ $$\int x^m(1-x)^ndx=\frac{x^n(1-x)^{m+1}}{m+1}+\frac{n}{m+1}\int x^{m+1}(1-x)^{n-1}dx$$ – bgins Apr 9 '12 at 13:44

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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Ah, I see the light at last... I think it might help if you explained first (for us mere mortals) that $$\eqalign{(1-x)^4&=x^4-4x^3+6x^2-4x+1\\&=(x^2+1-4x)(1+x^2)+4x^2}$$ $$\frac{(1-x)^4}{1+x^2}=x^2+1-4x+4\frac{x^2}{1+x^2}$$ $$\frac{x^4(1-x)^4}{1+x^2}=x^6+x^4-4x^5+4\frac{x^6}{1+x^2}$$ $$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^8+x^6}{1+x^2}+\frac{x^6+x^4}{1+x^2}-4\frac{x^‌​7+x^5}{1+x^2}+4\frac{x^6}{1+x^2}$$ – bgins Apr 9 '12 at 23:37
@bgins Do you have a proof that you are mortal or it is just your hypothesis.... – pedja Apr 10 '12 at 9:10
@pedja, LOL. Can I say its a fact? – Jiew Meng Apr 10 '12 at 12:25
@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. – bgins Apr 10 '12 at 13:34

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