William Lowell Putnam Integral Problem [closed]

Prove That $$\frac{22}{7}-\pi = \int_{0}^{1}\frac{x^{4}\left(1 - x\right)^{4}}{1 + x^{2}}\,{\rm d}x$$

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closed as off-topic by Carl Mummert, Chappers, graydad, Newb, Arctic CharApr 27 '15 at 4:37

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– Lucian Apr 16 '14 at 5:29
Thanks Lucian for the link – Ekaveera Kumar Sharma Apr 16 '14 at 6:31
I just add $\large{\rm d}x$. – Felix Marin May 15 '14 at 4:55

4 Answers

Literally all you have to do is expand it out, divide, and integrate.

$$x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$$ Now use long division: $$x^4(1-x)^4 = x^6(x^2+1)- 4x^5(x^2+1) + 5x^4(x^2 + 1) - 4x^2(x^2+1)+4(x^2+1) -4 \\ \frac{x^4(1-x)^4}{1+x^2} = (x^6 -4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{x^2+1} \\$$ We integrate this: $$\int_0^1 \! (x^6 -4x^5 + 5x^4 - 4x^2 + 4) \,\mathrm{d}x - \int_0^1 \! \frac{4}{x^2+1} \, \mathrm{d}x \\ = \frac{22}{7} - 4(\arctan(1) - \arctan(0)) \\ = \frac{22}{7} - \pi$$

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Crap, I just read noticed the Wikipedia link gave the exact same solution LOL – CosmoVibe Apr 16 '14 at 5:58

I Tried in this way:

$$\frac{x^4\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1+1\right)\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1\right)\,\left(1-x\right)^4}{1+x^2}+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x^2-1\right)\left(1-x\right)^4+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x+1\right)\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x-1\right)^6+2\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}$$ First two terms are straight forward to integrate.

$$\int_0^1 \frac{\left(1-x\right)^4}{1+x^2}=\int_0^1\left(x^2-4x+5\right)-\frac{4}{x^2+1}=\frac{10}{3}-\pi$$

Can any one suggest ant alternate ways to do this problem..

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Note that denominator is of lower degree than numerator. this is the only method I can think of. well done – evil999man Apr 16 '14 at 5:26

\begin{align}\int_0^1 dx \frac{x^4 (1-x)^4}{1+x^2} &= \int_0^1 dx \frac{x^4-4 x^5 + 6 x^6 - 4 x^7 + x^8}{1+x^2}\\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{4+2 k}-4 x^{5+2 k}+6 x^{6+2 k}-4 x^{7+2 k}+x^{8+2 k}\right )\\&= \sum_{k=0}^{\infty} (-1)^k \left (\frac1{5+2 k}-\frac{4}{6+2 k}+\frac{6}{7+2 k}-\frac{4}{8+2 k}+\frac1{9+2 k} \right ) \end{align}

Now,

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{5+2 k} = \frac15-\frac17+\frac19-\cdots=\frac{\pi}{4}-\left (1-\frac13 \right )$$

Similarly,

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{7+2 k} = \frac17-\frac19+\frac1{11}-\cdots=\left (1-\frac13+\frac15 \right ) -\frac{\pi}{4}$$ $$\sum_{k=0}^{\infty} \frac{(-1)^k}{9+2 k} = \frac19-\frac1{11}+\frac1{13}-\cdots=\frac{\pi}{4}-\left (1-\frac13+\frac15-\frac17 \right )$$

Meanwhile,

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{3+ k} = \frac13-\frac14+\frac15-\cdots=\log{2}-\left (1-\frac12 \right )$$ $$\sum_{k=0}^{\infty} \frac{(-1)^k}{4+ k} = \frac14-\frac15+\frac16-\cdots=\left (1-\frac12+\frac13 \right )-\log{2}$$

The result is, summing all of the above numbers,

$$-\left (1-\frac13 \right )+ 6 \left (1-\frac13+\frac15 \right )-\left (1-\frac13+\frac15-\frac17 \right )- \frac{2}{3} - \pi = \frac{22}{7}-\pi$$

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$\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx\;,$$ Let $\displaystyle x= \tan \phi\;,$ Then $dx = \sec^2 \phi.$

And Changing Limit, We Get $$\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^4 \phi\cdot \left(1-\tan \phi\right)^4d\phi.$$

$$\displaystyle =\int_{0}^{\frac{\pi}{4}}\tan^4 \phi \cdot \left(1-4\tan \phi+6\tan^2 \phi-4\tan^3 \phi+\tan^4 \phi\right).$$

$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\tan^4 \phi-4\tan^5 \phi+6\tan^6 \phi-4\tan^7 \phi+\tan^8 \phi \right\}d\phi.$$

$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\left[\tan^4 \phi+\tan^6 \phi \right]-4\left[\tan^5 \phi \cdot \sec^2 \phi \right]+\left[\tan^6 \phi+\tan^8 \phi \right]+4\tan^6 \phi \right\}d\phi.................(\star)\color{\red}\checkmark.$$

Now Let $$\displaystyle J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n}\phi d\phi\;,$$ Then $$\displaystyle J_{n+2} = \int_{0}^{\frac{\pi}{4}}\tan^{n+2}\phi d\phi$$

Now $$\displaystyle J_{n+2}+J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n} \phi \cdot \sec^2 \phi d\phi = \frac{1}{n+1}\Rightarrow J_{n+2}+J_{n} = \frac{1}{n+1}............(\star \star)\color{\red}\checkmark.$$

Now at $n=0\;,$ we get $\displaystyle J_{0} = \frac{\pi}{4}\;,$ Then $\displaystyle J_{2} = \left(1-\frac{\pi}{4}\right)$ and $\displaystyle$ When $n= 2\;,$ We get $$\displaystyle J_{4} = \left(\frac{\pi}{4}-\frac{2}{3}\right).$$

and When $n=4\;,$ We get $\displaystyle J_{6} = \left(\frac{13}{15}-\frac{\pi}{4}\right).$ and We get $\displaystyle J_{4}+J_{6}=\frac{1}{5}$ And $\displaystyle J_{6}+J_{8} = \frac{1}{7}$

Now Put these values in $............(\star)\color{\red}\checkmark.\;,$ We get

$$\displaystyle I = \frac{1}{5}-\frac{2}{3}+\frac{1}{7}+\frac{52}{15}-\pi = \left(\frac{22}{7}-\pi\right).$$

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