## Kainui one year ago I call this problem "the big kid's club": $\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx$

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1. anonymous

sweet baby jesus

2. IrishBoy123

.

3. anonymous

is this calculus

4. Kainui

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

5. Kainui

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

6. geerky42

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

7. 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

8. 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.

9. Kainui

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

10. Kainui

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

11. freckles

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

12. 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

13. Kainui

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

14. anonymous

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

15. anonymous

i knew it was something to gamma :-\

16. 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

17. dinamix

@Kainui we can use Integration by parts right ?

18. 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

19. freckles

i think n is a positive integer

20. freckles

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

21. dan815

i already know the solution so i wont comment

22. 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}

23. freckles

wow you made it look not so terrible

24. ganeshie8

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

25. freckles

works for n=2

26. anonymous

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

27. 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.

28. anonymous

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

29. 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 $$\large \color{black}{\sqrt{2 \pi n}}$$ .