How can I simplify $\lim_{n\to \infty } \, \int_0^n e^{-x} \left(1-\frac{x}{n}\right)^n \, dx$?

Question

How can I simplify $\lim_{n\to \infty } \, \int_0^n e^{-x} \left(1-\frac{x}{n}\right)^n \, dx$ ?

Limit[Integrate[(1 - x/n)^n/E^x, 
   {x, 0, n}], n -> Infinity]

The result was

Limit[ConditionalExpression[$\left.e^{-n} (-n)^{-n} (\Gamma (n+1,-n)-\Gamma (n+1)),\Re(n)>0\land \Im(n)=0\right],n\to \infty]$

Limit[ConditionalExpression[
   (-Gamma[1 + n] + Gamma[1 + n, 
      -n])/(E^n*(-n)^n), 
   Re[n] > 0 && Im[n] == 0], 
  n -> Infinity]

was supposed to be $1$ since

$\int_0^{\infty } e^{-x} \, dx=1$

Integrate[E^(-x), {x, 0, Infinity}]

I did try Simplify and Reduce but neither work.

Thank you in advance

Answer 1

$(1-\frac{x}{n})^n$ may be approximated by $exp(-\frac{x}{n})^n$

Exp[-1/n x] // Series[#, {x, 0, 3}] &

Limit[Exp[-1/n x]^n - (1 - x/n)^n, n -> Infinity]

=>

0

Limit[Integrate[Exp[-x]*Exp[-1/n x]^n, {x, 0, n}], n -> Infinity]

=>

1/2

---

About the error

Table[Exp[-1/n x]^n - (1 - x/n)^n, {n, 500, 10000, 500}] // 
 Plot[#, {x, 0, 100}] &

Answer 2

You can move the limit inside the integral (please see Rudin, Principles of Mathematical Analysis, Third Edition, Lebesgue's dominated convergence theorem, p. 321):

Limit[Integrate[Exp[-x] Limit[(1 - x/n)^n, n -> \[Infinity]],
      {x, 0, n}], n -> \[Infinity]]

Answer:

 1/2

As you surely know, $$\lim_{n\rightarrow\infty}\left(1-\frac{x}{n}\right)^n = e^{-x}$$

So the integral becomes

$$\int_0^\infty e^{-2x} dx=\frac{1}{2}$$

Answer 3

Workaround:

We have a functionE^(-a*x) (1 - x/n)^n where a=1 then:

func=E^(-a*x) (1 - x/n)^n;
sol1 = InverseLaplaceTransform[func, a, s]
(*(1 - x/n)^n DiracDelta[s - x]*)
sol2 = Integrate[sol1, {x, 0, n}, Assumptions -> {s > 0, n < Infinity}]
(*(1 - s/n)^n (DiscreteDelta[n] + HeavisideTheta[n - s] - HeavisideTheta[-n, n - s])*)
sol3 = Limit[sol2, n -> Infinity, Assumptions -> s > 0]
(*E^-s*)
sol4 = LaplaceTransform[sol3, s, a]
(*1/(1 + a)*)
Limit[sol4, a -> 1]
(*1/2*)

By numerics:

f[n_?NumericQ] := NIntegrate[(1 - x/n)^n*E^(-x), {x, 0, n}, Method -> "LocalAdaptive"];
Table[f[10^n], {n, 0, 6}]
(*{0.367879, 0.487198, 0.498747, 0.499875, 0.499987, 0.499999, 0.5} *)

Answer is 1/2 not 1.

Answer 4

Firstly, it is not entirely correct to approximate $\left(1-\frac{x}{n}\right)^n$ as $e^{-x}$ because even though $n$ goes to infinity, the integration probes the region $x\sim n$, hence there are contributions in the integral for which $\frac{x}{n}$ is not arbitrarily small!

Of course $e^{-x}$ suppresses all such contributions; nonetheless, IMHO it is not warranted to apply the limit inside the integral.

Let us instead consider the following. With the change of variables $x=n a$, we have $$\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{n}e^{-x}\left(1-\frac{x}{n}\right)^ndx=\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{1}ne^{-na}\left(1-a\right)^nda=\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{1}ne^{-na}\sum\limits_{k=0}^{\infty}(-1)^ka^k\binom{n}{k}da$$ Note that the Taylor expansion behaves well for all integration region as $0<a<1$. Thus $$\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{n}e^{-x}\left(1-\frac{x}{n}\right)^ndx=\sum\limits_{k=0}^{\infty}(-1)^k\lim\limits_{n\rightarrow\infty}n\binom{n}{k}\int\limits_{0}^{1}e^{-na}a^kda$$

We can do the integration (e.g. with Mathematica) to find $$\lim\limits_{n\rightarrow\infty}\int\limits_{0}^{n}e^{-x}\left(1-\frac{x}{n}\right)^ndx=\sum\limits_{k=0}^{\infty}(-1)^k\lim\limits_{n\rightarrow\infty}n^{-k}\binom{n}{k}\left(k!-\Gamma(k+1,n)\right)=\sum\limits_{k=0}^{\infty}(-1)^k$$

Therefore, the limit does not exist! (Thanks to @user64494 for pointing out my typo) However, we can always get the regularized result (a.k.a do analytical continuation from the region it does exist):

Sum[(-1)^k, {k, 0, \[Infinity]}, Regularization -> "Abel"]

gives 1/2.