Spooky Integrals

$$ These integrals can be solved fairly straightforward by using gamma function properties.

For $k=1$ , we have $$ $\int_0^\infty x^{-\frac{1}{4}}e^{-16x}\;dx=\frac{1}{8}\int_0^\infty u^{-\frac{1}{4}}e^{-u}\;du=\frac{1}{8}\Gamma\left(\frac{3}{4}\right),\quad\Rightarrow\quad u=16x.$ $$ For $k=2$ , we have $$ $\int_0^\infty x^{-\frac{1}{2}}e^{-81x}\;dx=\frac{1}{9}\int_0^\infty v^{-\frac{1}{2}}e^{-v}\;dv=\frac{1}{9}\Gamma\left(\frac{1}{2}\right),\quad\Rightarrow\quad v=81x.$ $$ For $k=3$ , we have $$ $\int_0^\infty x^{-\frac{3}{4}}e^{-256x}\;dx=\frac{1}{4}\int_0^\infty w^{-\frac{3}{4}}e^{-w}\;dw=\frac{1}{4}\Gamma\left(\frac{1}{4}\right),\quad\Rightarrow\quad w=256x.$ $$ Therefore $$ $\begin{aligned} \prod_{k=1}^3\left[\int_0^\infty x^{-\frac{k}{4}}e^{-(k+1)^4x}\;dx\right] &=\frac{1}{8}\Gamma\left(\frac{3}{4}\right)\cdot\frac{1}{9}\Gamma\left(\frac{1}{2}\right)\cdot\frac{1}{4}\Gamma\left(\frac{1}{4}\right)\\ &=\frac{1}{9}\Gamma\left(\frac{1}{2}\right)\cdot\frac{1}{8}\Gamma\left(\frac{3}{4}\right)\cdot\frac{1}{4}\Gamma\left(\frac{1}{4}\right).\\ \end{aligned}$ $$ Note that $$ : $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ $$ and from Euler's reflection formula , we have $$ $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}.$ $$ Hence, $\begin{aligned} \prod_{k=1}^3\left[\int_0^\infty x^{-\frac{k}{4}}e^{-(k+1)^4x}\;dx\right] &=\frac{1}{9}\sqrt{\pi}\cdot\frac{1}{8}\Gamma\left(1-\frac{1}{4}\right)\cdot\frac{1}{4}\Gamma\left(\frac{1}{4}\right)\\ &=\frac{1}{288}\sqrt{\pi}\cdot\frac{\pi}{\sin\left(\frac{1}{4}\pi\right)}\\ &=\frac{\sqrt{2}}{288}\pi^{\frac{3}{2}}. \end{aligned}$ $$ Thus, $a+b+c+d=2+288+3+2=\boxed{\color{#3D99F6}{295}}$ . $$

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

Alternate Solution:

Let, $I=\int _{ 0 }^{ \infty } x^{ -\frac { k }{ 4 } }e^{ -(k+1)^{ 4 }x }\; dx\\ =M\left\{ e^{ -(k+1)^{ 4 }x } \right\} \left( 1-\frac { k }{ 4 } \right)$

Here M{f(x)}(s) is Mellin transform of f(x) over s.

Now using Maclaurin series, $e^{ -(k+1)^{ 4 }x }=\sum _{ n=0 }^{ \infty }{ \frac { -(k+1)^{ 4n } }{ n! } { \left( -x \right) }^{ n } }$

Therefore, by Ramanujan's Master Theorem , $I={ \left( k+1 \right) }^{ k-4 }\Gamma \left( 1-\frac { k }{ 4 } \right)$

Now taking the product we get: $\prod _{ k=1 }^{ 3 } \left[ \int _{ 0 }^{ \infty } x^{ -\frac { k }{ 4 } }e^{ -(k+1)^{ 4 }x }\; dx \right] =\frac { \sqrt { 2 } }{ 288 } \pi ^{ \frac { 3 }{ 2 } }$

Gamma functions and Euler's Reflection Formula!!!!!