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An integral from maths world; pi formulas (50),

$$\pi=\frac{22}{7}-\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$

We found another similar integral to it, via experimental with wolfram integrator,

$$\pi=\frac{22}{7}-\frac{\pi}{4}\int_0^{\infty}\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx$$

Can anyone help us to prove it?

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    Wouldn't be $0=\frac{\pi}{4}\int_0^1\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx-\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$ the interesting part? – Patrick Abraham May 07 '16 at 20:16
  • This is not true - set $u = \exp(-\pi x/2)$ to get that the second integral is strictly less than $22/7 - \pi$. The same substitution also tells you that you'll get equality if you integrate over the positive reals. – stochasticboy321 May 07 '16 at 20:37
  • Correction on the limit on the second integrand –  May 07 '16 at 21:02

1 Answers1

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Note that $\displaystyle\cosh x=\frac{e^x+e^{-x}}{2}$. Let $\displaystyle t=e^{-\frac{x\pi}{2}}$, then $\displaystyle dt=-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx$. Also, we have $t=1$ as $x=0$, and $t\to0$ as $x\to\infty$. Therefore \begin{align} \frac{\pi}{4}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx &=\frac{\pi}{4}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\frac{e^{\frac{x\pi}{2}}+e^{-\frac{x\pi}{2}}}{2}}dx\\ &=\frac{\pi}{2}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{e^{\frac{x\pi}{2}}+e^{-\frac{x\pi}{2}}}\left(-\frac{2}{\pi}e^{\frac{x\pi}{2}}\right)\left(-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx\right)\\ &=-\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{1+e^{-x\pi}}\left(-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx\right)\\ &=-\int_1^0\frac{t^4\left( 1-t\right)^4}{1+t^2}dt\\ &=\int_0^1\frac{t^4\left( 1-t\right)^4}{1+t^2}dt. \end{align}

Solumilkyu
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