I Showing relationship between zeta and gamma

[howabout1337]

How can I show that $\int_0^\infty e^{-nt}\,t^{x-1}\,dt=\Gamma(x)\,n^{-x}\,$ for $x \geq 1\,$?

 

[Charles Link]

It would help if you included the equation $\Gamma(x)=\int\limits_{0}^{+\infty} e^{-t} \, t^{x-1} \, dt$ that I found in Wikipedia. I think all you need to do is let $t=n t'$.

 

[howabout1337]

It would help if you included the equation $\Gamma(x)=\int\limits_{0}^{+\infty} e^{-t} \, t^{x-1} \, dt$ that I found in Wikipedia. I think all you need to do is let $t=n t'$.

is the $t'$ in $t=n t'$ equivalent to $dt/dx$?

 

[Charles Link]

is the $t'$ in $t=n t'$ equivalent to $dt/dx$?

With $t=nt'$, you have $dt=n \, dt'$. (The Wikipedia article actually had $\Gamma(z)$, and their integration was done over $x$, but changing the letters doesn't change the mathematics). For the case at hand, $t$ is unrelated to $x$, and $dt/dx$ really has no meaning. (Note also, once you substitute in $t=nt'$, you are completely free to change the $t'$ back to the letter $t$).

 

[howabout1337]

With $t=nt'$, you have $dt=n \, dt'$. (The Wikipedia article actually had $\Gamma(z)$, and their integration was done over $x$, but changing the letters doesn't change the mathematics). For the case at hand, $t$ is unrelated to $x$, and $dt/dx$ really has no meaning. (Note also, once you substitute in $t=nt'$, you are completely free to change the $t'$ back to the letter $t$).

I ran through it, I got the equation. Does letting $t=nt'$ change the equation? Can I arbitrarily do this?

 

[Charles Link]

I ran through it, I got the equation. Does letting $t=nt'$ change the equation? Can I arbitrarily do this?

Basically you are showing that $\Gamma(x)=n^x \, \int\limits_{0}^{+\infty} e^{-nt} \, t^{x-1} \, dt$, and showing that the $n$ that appears on the right side of this last equation has no effect on the equation. It turns out, you can set $n$ equal to any number you want (Editing: It appears that $n$ must be positive or you have an $n^x$ that would be hard to define). $n$ is a constant, but that's all you know about $n$. The substitution $t=t'/n$ is completely legitimate, and then $dt= dt' /n$, and the limits of the integral are left unchanged. (The substitution is $t=t'/n$ if you begin with this expression that contains $n^x$).

 

[howabout1337]

Basically you are showing that $\Gamma(x)=n^x \, \int\limits_{0}^{+\infty} e^{-nt} \, t^{x-1} \, dt$, and showing that the $n$ that appears on the right side of this last equation has no effect on the equation. It turns out, you can set $n$ equal to any number you want (Editing: It appears that $n$ must be positive or you have an $n^x$ that would be hard to define). $n$ is a constant, but that's all you know about $n$. The substitution $t=t'/n$ is completely legitimate, and then $dt= dt' /n$, and the limits of the integral are left unchanged. (The substitution is $t=t'/n$ if you begin with this expression that contains $n^x$).

Does defining t as something else related to n such $t=t'/n^2$ or $t=2t'/n$ for any purpose changes the equation in different ways?

 

[Charles Link]

Does defining t as something else related to n such $t=t'/n^2$ or $t=2t'/n$ for any purpose changes the equation in different ways?

The answer is no, it doesn't change what it represents, but it makes it unnecessarily more complicated. Even in the original form that you have, the $n$ is unnecessary, and setting it equal to 1 simplifies the expression. $\\$ Editing: One other comment is that $n$ needs to be positive for a second reason: (It was previously pointed out that $n^x$ could become ambiguous if $n$ were negative). A second reason is that $e^{-nt}$ would be $e^{+|n|t}$, and the integral would diverge with the exponential becoming arbitrarily large.