Power of e

${ J }_{ 0 }=\int _{ -\infty }^{ 0 }{ { { x }^{ 0 }e }^{ nx }\quad dx } =\frac { 1 }{ n } \\ define\quad { J }_{ r }=\frac { d{ J }_{ r-1 } }{ dn } \\ \\ \frac { d{ J }_{ 0 } }{ dn } ={ J }_{ 1 }=\int _{ -\infty }^{ 0 }{ x{ e }^{ nx }\quad dx } =\quad \frac { -1 }{ { n }^{ 2 } } \\ \frac { d{ J }_{ 1 } }{ dn } ={ J }_{ 2 }=\int _{ -\infty }^{ 0 }{ { x }^{ 2 }{ e }^{ nx }\quad dx } =\quad \frac { 2 }{ { n }^{ 3 } } \\$

What we are doing here is differentiating the integral with respect to $n$ . Since we are differentiating with respect to $n$ , we can treat $x$ as a constant. And hence it can be seen that the $r^{th}$ derivative of ${J}_{0}$ will contain $r^{th}$ power of $x$ inside the integral.

$\\ \Rightarrow \frac { d{ J }_{ r-1 } }{ dn } ={ J }_{ r }=\int _{ -\infty }^{ 0 }{ { x }^{ r }{ e }^{ nx }\quad dx } =\quad \frac { { (-1) }^{ r }r! }{ { n }^{ r+1 } } \\ { \Rightarrow J }_{ r }=I(r,n)=\frac { { (-1) }^{ r }r! }{ { n }^{ r+1 } } \\ \Rightarrow { (I(r,n)) }^{ -1 }=\frac { { { (-1) }^{ r }n }^{ r+1 } }{ r! } \\ \Rightarrow \sum _{ r=0 }^{ \infty }{ { (I(r,n)) }^{ -1 } } =\quad { ne }^{ -n }\\ \Rightarrow \sum _{ n=1 }^{ \infty }{ \sum _{ r=0 }^{ \infty }{ { (I(r,n)) }^{ -1 } } } =\quad X\quad =\quad \frac { 1 }{ e } +\frac { 2 }{ { e }^{ 2 } } +\frac { 3 }{ { e }^{ 3 } } +\frac { 4 }{ { e }^{ 4 } } +...\\ \quad X=\frac { 1 }{ e } +\frac { 2 }{ { e }^{ 2 } } +\frac { 3 }{ { e }^{ 3 } } +\frac { 4 }{ { e }^{ 4 } } +...\\ \frac { X }{ e } =\quad \quad \quad \frac { 1 }{ { e }^{ 2 } } +\frac { 2 }{ { e }^{ 3 } } +\frac { 3 }{ { e }^{ 4 } } +...\\ \Rightarrow X(1-\frac { 1 }{ e } )=\frac { 1 }{ e } +\frac { 1 }{ { e }^{ 2 } } +\frac { 1 }{ { e }^{ 3 } } +\frac { 1 }{ { e }^{ 4 } } +...=\quad \frac { 1 }{ e-1 } \\ \Rightarrow X=\frac { { e } }{ { (e-1) }^{ 2 } } =0.92067\\ \Rightarrow \left\lfloor { 10 }^{ 4 }X \right\rfloor =\boxed { 9206 }$

Well, Raghav's solution is interesting but here is mine just for diversity.

$\displaystyle I(r,n) = \int_{-\infty}^{0}{{x}^{r}{e}^{nx} dx} = -\int_{0}^{\infty}{{x}^{r}{e}^{nx} dx}$

Substitute $\displaystyle nx = -u \Rightarrow x = \frac{-u}{n} \Rightarrow \frac{-du}{n} = dx$

$\displaystyle = -\int_{0}^{\infty}{\frac{{(-1)}^{r}{u}^{r}}{{n}^{r}}{e}^{-u}\frac{-du}{n}}$

$\displaystyle = \frac{{(-1)}^{r}}{{n}^{r+1}}\int_{0}^{\infty}{{u}^{r}{e}^{-u} du}$

$\displaystyle = \frac{{(-1)}^{r}}{{n}^{r+1}}*\Gamma(r+1)$

$\displaystyle \frac{1}{I(r.n)} = n\frac{{(-n)}^{r}}{r!}$

$\displaystyle \sum_{n=1}^{\infty}{n}\sum_{r=0}^{\infty}{\frac{{(-n)}^{r}}{r!}}$

Well, the sum is quite easy as we can see that the first sum is just equal to ${e}^{-n}$ . Then the final sum becomes -

$\displaystyle \sum_{n=1}^{\infty}{n{e}^{-n}}$

This can be solved easily by considering this base sum - $\displaystyle 1 + x + {x}^{2} +... = \frac{1}{1-x}$ (first differentiating this, then multiplying it by $x$ and then substituting $x = e$ .

Therefore, we get the final answer as -

$\displaystyle \frac{e}{{(e-1)}^{2}}$

$\begin{aligned} I(r,n) & = \int_{-\infty}^0 x^r e^{nx} dx & \small {\color{#3D99F6}\text{Let }-t = nx, \ x = -\frac tn, \ dx = -\frac 1n \ dt} \\ & = \int_\infty ^0 \left(-\frac tn \right)^r e^{-t} \left(-\frac 1n\right) dt \\ & = \frac {(-1)^r {\color{#3D99F6}\Gamma (r+1)}}{n^{r+1}} & \small {\color{#3D99F6}\Gamma(\cdot) \text{ is Gamma function}} \\ & = \frac {(-1)^r n!}{n^{r+1}} \end{aligned}$

Therefore, we have:

$\begin{aligned} X & = \sum_{n=1}^\infty \sum_{r=0}^\infty \frac 1{I(r,n)} \\ & = \sum_{n=1}^\infty \sum_{r=0}^\infty \frac {(-1)^r n^{r+1}}{n!} \\ & = \sum_{n=1}^\infty n \sum_{r=0}^\infty \frac {(-1)^r n^r}{n!} \\ & = \sum_{n=1}^\infty n e^{-n} & \small {\color{#3D99F6} \text{Note that } \sum_{n=1} ^\infty n e^{-n}= \sum_{n=0}^\infty e^{-n}} \\ & = \sum_{\color{#3D99F6}n=0}^\infty \frac {\color{#3D99F6}n+1}{e^{\color{#3D99F6}n+1}} \\ & = \frac 1e {\color{#3D99F6} \sum_{n=0}^\infty \frac n{e^n}} + \frac 1e \sum_{n=0}^\infty \frac 1{e^n} \\ & = \frac {\color{#3D99F6} X}e + \frac 1e \left(\frac 1{1-\frac 1e} \right) \\ \left(1-\frac 1e \right) X & = \frac 1e \left(\frac e{e-1} \right) \\ \implies X & = \frac e{(e-1)^2} \approx 0.920673594 \end{aligned}$

$\implies \left \lfloor 10^4X \right \rfloor = \boxed{9206}$

gamma function ,

then taylor series if e,

then diff w.r.t x of sum e^(-x)