Question

I call this problem "the big kid's club":

\[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]

Answer 1

sweet baby jesus

Answer 2

.

Answer 3

is this calculus

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Yeah, I did actually, specifically the formula in here they have for the integral to n!

\[n! = \int_0^\infty x^ne^{-x}dx\]

Answer 14

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

Answer 15

i knew it was something to gamma :-\

Answer 16

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

Answer 17

@Kainui we can use Integration by parts right ?

Answer 18

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

Answer 19

i think n is a positive integer

Answer 20

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

Answer 21

i already know the solution so i wont comment

Answer 22

let \(x=-u \) for the first integral,

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

Answer 23

wow you made it look not so terrible

Answer 24

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

Answer 25

works for n=2

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

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 \(\large \color{black}{\sqrt{2 \pi n}}\) .