I call this problem "the big kid's club":
\[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]

- Kainui

- jamiebookeater

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- anonymous

sweet baby jesus

- IrishBoy123

.

- anonymous

is this calculus

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## More answers

- Kainui

WHOOPS! Slight mistake in the signs ahehehe...
\[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]

- Kainui

I just went through it again to make sure I didn't make any mistakes in coming up with this thing.

- geerky42

Binomial theorem? I don't think it is hard, but it is very tedious...

- freckles

\[\int\limits_{0}^{n}[(1+\frac{x}{n})^n e^{-x}+(1-\frac{x}{n})^n e^{x}] dx+\int_n^\infty (1+\frac{x}{n})^n e^{-x}dx\]
I wonder if we can do something with this

- Kainui

I didn't use the binomial theorem to solve this, but you're welcome to try. I guess I should also mention that you can keep n as a positive integer if you like, but if you solve it more generally I will be impressed cause I don't know how to do that.

- Kainui

@freckles That looks like it might be promising. :D

- Kainui

Well... It's on the right track, considering the bounds of the integrals, I'll say that much.

- freckles

did you get this from playing around with some stirling type stuff?

- freckles

\[\int\limits_0^n n^{-n}e^{-x}(e^{2x}(n-x)^n+(n+x)^n) dx+\int\limits_n^\infty (1+\frac{x}{n})^n e^{-x}\]
like some of the factors in the first integrand look similar to some factors in integrand on that one stirling page

- Kainui

Yeah, I did actually, specifically the formula in here they have for the integral to n!
\[n! = \int_0^\infty x^ne^{-x}dx\]

- anonymous

So what language is this called I know its not English...

- anonymous

i knew it was something to gamma :-\

- Kainui

Haha. I'm going to go down and swim in the lake since it's a beautiful day, in about 30 minutes to an hour I'll be back to give the answer I think or hints... or hopefully even congratulate whoever gets it :P

- dinamix

@Kainui we can use Integration by parts right ?

- anonymous

i am able to reduce it to one integral.. is there any constraint to 'n' coz if the limit tends to infinite then um getting 1^infinite form

- freckles

i think n is a positive integer

- freckles

though he said something about if you find a formula for n being a negative integer he would be really impressed

- dan815

i already know the solution so i wont comment

- ganeshie8

let \(x=-u \) for the first integral,
\[\begin{align}
&\int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\
&=\int_{-\infty}^0 (1-\frac{u}{n})^n e^{u}du + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\
&=\int_{-\infty}^n (1-\frac{x}{n})^n e^{x}dx \\~\\
&=ne^n\int_0^{\infty}t^n e^{-nt}dt \\~\\
&=e^n\dfrac{\Gamma(n+1)}{n^n}
\end{align}\]

- freckles

wow you made it look not so terrible

- ganeshie8

Haha I hope it is right, didn't verify if it really works...

- freckles

works for n=2

- anonymous

@ganeshie8 it is beautiful and not crazy!! :)

- Kainui

Yeah that's exactly right! This is why I called it "the big kid's club" because
\[\huge \frac{e^n n!}{n^n}\]
\(e^n\), \(n!\), \(n^n\) are 3 popular big functions, all in one place haha.

- anonymous

@Kainui Brilliant i was waiting to hear why u called it that more than the way to solve xD

- SolomonZelman

Just a little addition, that according to Stirling's approximation,
\(\large\color{black}{ \displaystyle n! {\LARGE\text{ ~}}\sqrt{2\pi n}\frac{n^n}{e^n} }\)
So, when we apply that:
\(\large\color{black}{ \displaystyle \frac{\color{red}{n!}e^n}{n^n} =}\)\(\large\color{black}{ \displaystyle \frac{\color{red}{\sqrt{2\pi n}\frac{n^n}{e^n}}e^n}{n^n}= }\)\(\large\color{black}{ \displaystyle \frac{\color{red}{\sqrt{2\pi n}~n^n}}{n^n} =}\)\(\large\color{black}{ \displaystyle \sqrt{2\pi n}}\)
But, considering the fact that Factorial is greater than its Stirling's approximation
of it, (for all whole number x), your value is really a little greater than \(\large \color{black}{\sqrt{2 \pi n}}\) .

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