compute: $\int_{0}^{1}e^x(1-x)^{100}dx$ I'm trying to compute: $\int_{0}^{1}e^x(1-x)^{100}dx$. I tried to use integration by parts but it didn't work out for me(since I need to do that 100 times, and obviously there's a shorter solution) ,  I substituted  $(1-x)=u$ and got $e\int_0^1e^{-t}t^{100}$, again I can't do with that much. Any suggestion how should solve this integral?
Thanks a lot guys!
 A: The integral is exactly the fractional part of $100!\,e$, or in other words $100!\ e-\lfloor100!\ e\rfloor\approx0.00999901019\ldots$
Apply integration by parts to the integral $I_n=\int_0^1e^{1-t}t^n\,dt$ (it's nicer not to pull the $e$ out to the front) and we find for $n\geq1$, $$I_n=-1+nI_{n-1}$$
This gives us $$I_{100}=-1+100[-1+99[-1+98[-1+\cdots+2[-1+1I_0]\cdots]]]$$
$I_0$ is a straightforward computation: $e-1$. So 
This gives us $$I_{100}=-1+100[-1+99[-1+98[-1+\cdots+2[-1+e-1]\cdots]]]$$
Here is a nice observation. Once this is multiplied out, it (clearly?) simplifies to $100!\,e-N$ for some integer $N$. A graphical examination of the integral reveals that $I_{100}$ is somewhere between $0$ and $1$. (You could prove this using the fact that $e^{1-t}t^{100}=e^{1-t}tt^{99}\leq t^{99}$ on $[0,1]$.) So $N$ must equal the integer part of $100!\,e$, leaving $I_{100}$ to be the fractional part.

It's interesting to note that since $I_n\to0$ as $n\to\infty$, the fractional part of $n!\,e$ must approach zero; that is, $n!\,e$ gets closer and closer to being an integer. (Although I suppose that is obvious if we consider the usual series expansion for $e$.)

For computational purposes, we can use this to find a decimal approximation by throwing out the first $100$ terms or so (which are all integers) of the series expansion for $100!\, e$.
$$
\begin{align}
\int_0^1e^{1-t}t^{100}\,dt & 
= 100!\, e-\lfloor100!\,e\rfloor\\
& = \sum_{n=101}^{\infty}\frac{100!}{n!}
\end{align}
$$
This is the series that bgins has found with a slightly different argument. At first, this series converges faster than David Mitra's alternating series. It is correct to at least 17 decimal places after only 8 partial summands. David's requires 18 partial summands to get that much accuracy. However since both series have a ratio of order $1/n$ and David's series is alternating, I think that in the long run for very high accuracy demands, his series might be better.
A: Use the formula 
$$
\frac{d}{dx}\left(e^x\ \sum_{n\ge0}\ (-1)^n\ f^{(n)}(x)\right)=e^x\ f(x),
$$
which holds if $f$ is a polynomial.
A: You could use integration by parts. You will end up with a recursion, like this:
$$\int\limits_0^1 {{e^{ - x}}{x^n}dx = \left[ { - {e^{ - x}}{x^n}} \right]_0^1}  + n\int\limits_0^1 {{e^{ - x}}} {x^{n - 1}}dx$$
$${I_n} =   -\frac{1}{e} + n{I_{n - 1}}$$
where 
$$\int\limits_0^1 {{e^{ - t}}{t^n}dt}  = {I_n}$$
Using it sufficient times you'll end up with your result. 
A: Substitute $1-x = t$ , so :
$I= -e \int \limits_1^0 e^{-t} \cdot t^{100}\,dt =e \int \limits_0^1 e^{-t} \cdot t^{100}\,dt$
Now substitute : $-t=s$ , so :
$I= -e \int \limits_0^{-1} e^{s} \cdot s^{100}\,ds = e \int \limits_{-1}^{0} e^{s} \cdot s^{100}\,ds$
This integral you can solve using Integration by parts several times .For start choose :
$u = s^{100}~$ , and $dv = e^s ds$
A: As already said, this is $I_n$ for $n=100$, where
$$
I_n=\int_0^1x^n\mathrm e^{1-x}\mathrm dx.
$$
To compute $I_n$ for every $n\geqslant0$ at once, one can group these in a series, that is, consider, for every $s$ small enough,
$$
\sum_{n\geqslant0}\frac{s^n}{n!}I_n=\int_0^1\sum_{n\geqslant0}\frac{s^n}{n!}x^n\mathrm e^{1-x}\mathrm dx=\int_0^1\mathrm e^{1-x(1-s)}\mathrm dx=\frac{\mathrm e-\mathrm e^{s}}{1-s}.
$$
The expansion of the first part of the RHS is 
$$
\frac{\mathrm e}{1-s}=\sum\limits_{n\geqslant0}\mathrm es^n.
$$
As regards the second part, one knows that
$$
\mathrm e^s\cdot\frac1{1-s}=\left(\sum\limits_{n\geqslant0}\frac{s^n}{n!}\right)\cdot\left(\sum\limits_{n\geqslant0}s^n\right),
$$
whose coefficient of $s^n$ is
$$
\sum\limits_{k=0}^n\frac1{k!}.
$$
Putting all these together yields
$$
\frac{I_n}{n!}=\mathrm e-\sum\limits_{k=0}^n\frac1{k!}=\sum\limits_{k=0}^{+\infty}\frac1{k!}-\sum\limits_{k=0}^n\frac1{k!}=\sum\limits_{k\geqslant n+1}\frac1{k!},
$$
and finally,
$$
I_n=\sum\limits_{k\geqslant1}\frac{n!}{(n+k)!},
$$
To get an approximate value of $I_n$, one can proceed as follows. Keeping only the first term of the series in the RHS yields a lower bound. Replacing the $k$th term by $\frac1{(n+1)^k}$ and summing the resulting series yields an upper bound. These read
$$
\frac1{n+1}\lt I_n\lt\frac1n.
$$
A: For what it's worth:
Write
$$
t^{100}e^{-t}=t^{100}(1-t+{t^2\over 2!}-{t^3\over 3!}-\cdots )
=t^{100}-t^{101}+{t^{102}\over 2!}-{t^{103}\over 3!}-\cdots
$$
The above series is uniformly convergent on $[0,1]$; thus:
$$
\eqalign{
\int_0^1 e^{-t}t^{100}\,dt
&=\sum_{n=0}^\infty \int_0^1 (-1)^n{t^{100+n}\over n!}\cr

&=\sum_{n=0}^\infty   (-1)^n{t^{101+n}\over({101+n}) n!}\Bigl|_0^1\cr
 
&=\sum_{n=0}^\infty   (-1)^n{1\over({101+n}) n!}. \cr
}
$$
A: For $n>-1$, consider $$I=\int\limits_{0}^{1}{\rm d}t\,e^{t}(1-t)^{n}.$$ With the substitution $u=1-t$ $$I=e \int\limits_{0}^{1}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}=e \Bigg(\int\limits_{0}^{\infty}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}-\int\limits_{1}^{\infty}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}\Bigg)=e\big[\Gamma(n+1)-\Gamma(n+1,1)\big],$$ where $\Gamma(s)$ and $\Gamma(s,a)$ are the gamma and incomplete gamma functions respectively. For $n=100$ we have $e\big[\Gamma(101)-\Gamma(101,1)\big]\approx 0.009999$.
A: $\int_0^1e^x(1-x)^{100}~dx$
$=\int_1^0e^{1-x}x^{100}~d(1-x)$
$=\int_0^1x^{100}e^{1-x}~dx$
$=-\left[\sum\limits_{n=0}^{100}\dfrac{100!x^ne^{1-x}}{n!}\right]_0^1$ (can be obtained from https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_of_polynomials)
$=100!e-\sum\limits_{n=0}^{100}\dfrac{100!}{n!}$
