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howabout1337
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How can I show that ##\int_0^\infty e^{-nt}\,t^{x-1}\,dt=\Gamma(x)\,n^{-x}\,## for ##x \geq 1\,##?
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is the ##t'## in ##t=n t'## equivalent to ##dt/dx##?Charles Link said: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' ##.
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 said:is the ##t'## in ##t=n t'## equivalent to ##dt/dx##?
I ran through it, I got the equation. Does letting ## t=nt' ## change the equation? Can I arbitrarily do this?Charles Link said: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 ##).
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 said:I ran through it, I got the equation. Does letting ## t=nt' ## change the equation? Can I arbitrarily do this?
Charles Link said: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 ##).
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.howabout1337 said: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?
Zeta (ζ) and gamma (γ) are related in the context of complex numbers. Zeta is the real part and gamma is the imaginary part of a complex number. They are also related in terms of the Riemann zeta function and the gamma function.
The Riemann zeta function is a mathematical function denoted by ζ(s) that is defined for all complex numbers s with a real part greater than 1. It is an important function in number theory and has connections to the distribution of prime numbers.
The Riemann zeta function and the gamma function are related through the functional equation ζ(s) = 2sπs-1sin(πs/2)Γ(1-s) where Γ is the gamma function. This relationship allows for the extension of the Riemann zeta function to all complex numbers, not just those with a real part greater than 1.
The relationship between zeta and gamma has many important implications in mathematics. It is used in number theory, complex analysis, and other areas of mathematics. For example, the Riemann zeta function is closely related to the distribution of prime numbers, and the gamma function is used in many mathematical models and calculations.
While the relationship between zeta and gamma may seem abstract, it has real-world applications in fields such as physics, engineering, and economics. For example, the Riemann zeta function is used in quantum field theory, and the gamma function is used in calculating probabilities in statistics and finance.