Ramanujan's Beautiful Master Formula: Integral From 0 to ∞ (e)^(-x^n)sin(x^n) dx

\begin{align*}& f(x)=\sum_{n=0}^{\infty} \phi(n) \dfrac{(-x)^n}{n!} \Longrightarrow
M \left[ f(s) \right] = \int_0^{\infty} x^{s-1}f(x) dx = \Gamma(s) \phi(-s) \\& \\&\therefore \int_0^ {\infty} e^{{-x}^n} \sin (x^n) dx = \dfrac{\left( \dfrac{1}{n} \right)! \sin \left( \dfrac{ \pi }{4n} \right)}{ \sqrt[2n]{2}} \\& \\&{\rm{Proof}}: \\& \\& \therefore u= {-x}^n \Longrightarrow I_n = \dfrac{1}{n} \int_0^{\infty} u^{\frac{1-n}{n}} e^{-u} \sin{u} \: du \\& \\&Note: \dfrac{1}{n} \int_0^{\infty} u^{\frac{1-n}{n}} e^{-u} \sin{u} =
\dfrac{1}{n} \Im \left[ \int_0^{\infty} u^{\frac{1-n}{n}} e^{-u+iu} \: du \right] \\& \\&Note: e^{-u+iu} = \sum_{n=0}^{\infty} \left( 2^{0.5n} e^{ -0.25in \pi} \right) \dfrac{(-x)^n}{n!} \\& \\&Note: \dfrac{1-n}{n}=s-1 \Longrightarrow s= \dfrac{1}{n} \\& \\&\therefore \dfrac{1}{n} \Im \left[ \int_0^{\infty} u^{\frac{1-n}{n}} e^{-u+iu} \: du \right] =\Im \left[ \Gamma \left( 1+ \dfrac{1}{n} \right) \dfrac{1}{ \sqrt[2n]{2}} e^{ \frac{i \pi}{4n}} \right] \\& \\&= \dfrac{\left( \dfrac{1}{n} \right)! \sin \left( \dfrac{ \pi }{4n} \right)}{ \sqrt[2n]{2}} \Box

For when n = 3 refer to Dr. Peyam's video below:

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.