Question

$\displaystyle \sf \sum \limits^{ \infty}_{a = 2} \sum \limits^{ \infty}_{b = 1} \int ^{ \infty}_{0} \frac{ {x}^{b} }{( {e}^{ax} )(b!)} \: dx$​

Answer 1

$\huge\orange{\frac{\pi^2}{6}+1}$

Step-by-step explanation:

First let's take integral part

$\int ^{ \infty}_{0} \frac{ {x}^{b} }{( {e}^{ax} )(b!)} \: dx \\ \\ put \: ax = y \: \implies \: dx = \frac{dy}{a} \\ \\ \int ^{ \infty}_{0} \frac{ {( \frac{y}{a}) }^{b} }{( {e}^{y} )(b!)} \: \frac{dy}{a} \\ \\ = \frac{1}{ {a}^{b + 1}.b! } \int ^{ \infty}_{0} {y}^{b} {e}^{ - y} dy \\ \\ using \: gamma \: function \\ \\ = \frac{(b + 1)!}{ {a}^{b + 1}. b!} \\ \\ = \frac{b + 1}{ {a}^{b + 1} }$

Now our question is reduced to

$\displaystyle \sf \sum \limits^{ \infty}_{a = 2} \sum \limits^{ \infty}_{b = 1} \frac{b + 1}{ {a}^{b + 1} }$

Let

$let \: z = \frac{1}{a} < 1 \: then \:series \\ \\ s = z + {z}^{2} + ... = \frac{ z }{1 - z} \\ \\ differentiate \: wrt \: z \\ \\ 1 + 2z + 3 {z}^{2} + .... = \frac{(1 - z) + z}{(1 - z) {}^{2} } \\ \\ multiply \: both \: sides \: with \: z \\ \\ z + 2 {z}^{2} + 3 {z}^{3} + .... = \frac{z}{(1 - z) {}^{2} } \\ \\ put \: z = \frac{1}{a} \\ \\ \frac{1}{a} + \frac{2}{ {a}^{2} } + \frac{3}{ {a}^{3} } + .... = \frac{( \frac{1}{a} )}{(1 - \frac{1}{a}) {}^{2} } = \frac{ {a}^{2} }{(a - 1) {}^{2} } \\ \implies \\ \frac{2}{a {}^{2} } + \frac{3}{ {a}^{3} } + .... = \frac{ {a}^{2} }{(a - 1) {}^{2} } - \frac{1}{a} \\ \implies \\ \\ \sum \limits^{ \infty}_{b = 1} \frac{b + 1}{ {a}^{b + 1} } = \frac{a{}^{2} - {a}^{2} + 2a - 1 }{a(a - 1) {}^{2} }$

Put the value in summation we left with

$\\ \\ = \sum \limits^{ \infty}_{a = 2} \frac{ 2a - 1}{ {a}(a-1)^2}$

$\red{Rest \: solution \: is\: in \:picture}$