Deranged integral

$I\left( n \right)=\int _{ 0 }^{ \infty }{ { (t-1) }^{ n } } { e }^{ -t }dt$

We know from the binomial expansion of ${ (t-1) }^{ n }$ that:

${ (t-1) }^{ n }={ t }^{ n }-\left( \begin{matrix} n \\ 1 \end{matrix} \right) { t }^{ n-1 }+....+\left( \begin{matrix} n \\ r \end{matrix} \right) { t }^{ n-r }{ (-1) }^{ r }+....\left( \begin{matrix} n \\ n \end{matrix} \right) { (-1) }^{ n }\\ \quad \quad \quad \quad =\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } { t }^{ n-r }{ (-1) }^{ r }$

Inputting this in the given integral and exchanging integral with summation:

$I\left( n \right)=\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } { (-1) }^{ r }\int _{ 0 }^{ \infty }{ { t }^{ n-r } } { e }^{ -t }dt$

Now, invoking the gamma integral, we get:

$I\left( n \right) =\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } { (-1) }^{ r }\int _{ 0 }^{ \infty }{ { t }^{ n-r } } { e }^{ -t }dt\\ I\left( n \right) =\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } { (-1) }^{ r }(n-r)!\\ I\left( n \right) =\sum _{ r=0 }^{ n }{ \frac { n! }{ r!(n-r)! } } { (-1) }^{ r }(n-r)!\\ I\left( n \right) =n!\sum _{ r=0 }^{ n }{ \frac { { (-1) }^{ r } }{ r! } }$

This expression for $I\left( n \right)$ is same as the number of derangements for $n$ objects.

So,

$I\left( n \right) =n!\sum _{ r=0 }^{ n }{ \frac { { (-1) }^{ r } }{ r! } } =D\left( n \right)\\ I\left( n \right)-D\left( n \right)=0$

Cheers! :)

D(42)=[42!/e].

Multiply and divide I(n) by e, then we have

I(n) = (1/e)*G(n+1) = n!/e

Then I(42)=42!/e

{G represents the gamma function}

So I(42) - D(42) = 42!/e - [42!/e], which

is pretty much 0.