Integral definition of factorial

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Runei
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I'm watching V. Balakrishnan's video lectures on Classical Physics, and right now he's going through statistical mechanics.

In that regards he's talking about Stirlings formula, and at one point, he wrote an integral definition of the factorial like the following

[itex]n! = \int_{0}^{\infty}dx\hspace{0.1cm}e^{-x}\hspace{0.1cm}x^n\hspace{0.1cm},\hspace{2cm} \text{where}\hspace{1cm} n={1,2,3 ...}[/itex]

Why is he writing the integral in that way? With the dx first and the exponentials afterwards?

I thought the definition was

[itex]n! = \int_{0}^{\infty}e^{-x}x^ndx[/itex]

Can anybody explain this?

Many thanks in advance :)
 
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Ah, I was wondering that that might be the case :)

Many thanks Micromass!
 
Multiplication is associative! :P
 
If you think of ##\displaystyle\int_0^\infty \cdots \,dx## as an operator, writing ##\displaystyle\int_0^\infty\!\! dx\,f(x)## is just as sensible as something like ##\displaystyle\frac{d^n}{dx^n}f(x)## - or even ##\displaystyle\left(\frac 1 r \frac{d}{dr}\right)^nf(r)##.
 
While I perfectly understand the idea of the notation, i think its confusing in the way where the following two ways are different:

[itex]\int(dx x) = \int x dx = 1/2 x^2[/itex]

And

[itex]\int(dx) x = x^2[/itex]

(Mind the constants not shows)
 
Runei said:
[itex]\int(dx) x = x^2[/itex]

OK, but this is bad notation. The ##x## in the ##dx## is the dummy variable. The other ##x## is an actual variable. You shouldn't use the same name for them.
And indefinite integrals are tricky things so I'm not sure if the thing above is well-defined. See https://www.physicsforums.com/blog.php?b=4566
 
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micromass said:
But it's commutative we want! :-p

I fail
 
Runei said:
While I perfectly understand the idea of the notation, i think its confusing in the way where the following two ways are different:

[itex]\int(dx x) = \int x dx = 1/2 x^2[/itex]

And

[itex]\int(dx) x = x^2[/itex]

(Mind the constants not shows)
No this is wrong. The second form should be [itex]\left(\int dx\right)x= x^2[/itex].
 
micromass said:
It's the same thing. But physicists tend to use the notation of writing ##dx## first for some reason. It's just a notational issue. For all intents and purposes, we have

[tex]\int f(x)dx = \int dx f(x)[/tex]

HallsofIvy said:
We can't really expect physicists to write mathematics correctly, can we?:devil:

It's worth pointing out that mathematicians do this as well, specifically in papers dealing with n-fold integration. Cauchy's formula for repeated integration is very often written
##\int_{a}^{x} dt_n \int_{a}^{t_n} dt_{n-1} \ldots \int_{a}^{t_2} f(t_1) \, dt_1 = \frac{1}{(n-1)!}\int_a^x (x-t)^{n-1}\,f(t)\,dt##
Observe this uses both notations mixed!