The first published statement of this result was in 1971 by Dalzell. It was also presented as a problem in the 1968 Putnam Competition, and later it also came in the JEE-Adv. The problem was to compute the integral,

∫01x4(1−x)41+x2 dx.\int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx.

I encourage readers to try it at first.

Solution:

∫01x4(1−x)41+x2 dx=∫01x6−4x5+5x4−4x2+4−41+x2 dx=[x77−2x63+x5−4x33+4x−4arctan⁡x]01=17−23+1−43+4−π=227−π.\begin{aligned} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx &= \int_0^1 x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1 + x^2} \:dx \\ &= \left[\frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} + 4x - 4\arctan{x}\right]_0^1 \\ &= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - \pi \\ &= \frac{22}{7} - \pi. \end{aligned}

Now observe one thing:

x4(1−x)41+x2>0  ⟹  ∫01x4(1−x)41+x2 dx>0.\frac{x^4(1 - x)^4}{1 + x^2} > 0 \implies \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx > 0.

Hence,

227>π.\frac{22}{7} > \pi.


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About the author

Sougata Panda is a student of the Chennai Mathematical Institute. This particular article placed third in our Article Writing Contest, 2022.