KawaiiAmber

Hello! My name is Zack. I'm a huge weeb. haha! I like doing math stuff and watching anime. I like kiss anime as they seem to have the largest variety. Here are some handy notes that I've written up on some topics I like. Look for the appropriate section.

It has been mentioned that the gamma function was able to fill in the place of the factorial. The motivating logic behind this is mainly the Integration by parts rule. That is:

${\displaystyle \int \limits _{C}\!u\,\mathrm {d} v=\left.uv\right|_{C}-\int \limits _{C}\!v\,\mathrm {d} u}$ 

for some u and v of x over some boundary C. Consider the integral representation of the factorial of s:

${\displaystyle s!=\int \limits _{0}^{\infty }\!t^{s}e^{-t}\,\mathrm {d} t}$ 

Apply the Integration by parts rule for s = 3:

${\displaystyle 3!=\int \limits _{0}^{\infty }\!t^{3}e^{-t}\,\mathrm {d} t}$ 

In the case of factorials, one generally sets the power term to u.

${\displaystyle u=t^{3}}$ 

${\displaystyle \mathrm {d} v=e^{-t}\implies v=-e^{-t}}$ 

One can then apply Integration by parts:

${\displaystyle \int \limits _{0}^{\infty }\!t^{3}e^{-t}\,\mathrm {d} t=\left.-t^{3}e^{-t}\right|_{0}^{\infty }-\int \limits _{0}^{\infty }\!-e^{-t}3t^{2}\,\mathrm {d} t=\left.-t^{3}e^{-t}\right|_{0}^{\infty }+3\int \limits _{0}^{\infty }\!e^{-t}t^{2}\,\mathrm {d} t}$ 

Observe how it became 3 times the factorial of 2. Take the limits of the left side:

${\displaystyle \left.-t^{3}e^{-t}\right|_{0}^{\infty }=\left.{\frac {-t^{3}}{e^{t}}}\right|_{0}^{\infty }}$ 

One can use L'Hôpital's rule to show that this approaches zero. This implies that:

${\displaystyle 3!=3\int \limits _{0}^{\infty }\!t^{2}e^{-t}\,\mathrm {d} t}$ 

Regarding the limits of:

${\displaystyle \left.{\frac {-t^{n}}{e^{t}}}\right|_{0}^{\infty },}$ 

the power always reaches 0 by derivation before the exponential. For any n in this limit, it is shown to go to zero via L'Hôpital's rule. Regarding the integral part on the right, if one were to work out the second part for 3!, they would get:

${\displaystyle 3!=3\cdot 2\int \limits _{0}^{\infty }\!te^{-t}\,\mathrm {d} t}$ 

Integrating again yields:

${\displaystyle 3!=3\cdot 2\cdot 1\int \limits _{0}^{\infty }\!e^{-t}\mathrm {d} t=3\cdot 2\cdot 1\cdot 1}$ 

In general, it is seen that the integral on the right produces the factorial of a number that is one smaller multiplied by the number one is taking the factorial of in the first place. For every step in integration via L'Hôpital's rule, the limit regarding the power and exponential terms is shown to go to zero. The logic of Integration by parts shows that one always gets this cascading product on the right side. That is:

${\displaystyle \int \limits _{0}^{\infty }\!t^{n}e^{-t}\,\mathrm {d} t=\left.{\frac {-t^{n}}{e^{t}}}\right|_{0}^{\infty }+n\int \limits _{0}^{\infty }\!t^{n-1}e^{-t}\,\mathrm {d} t}$ 

${\displaystyle \left.{\frac {-t^{n}}{e^{t}}}\right|_{0}^{\infty }=0\implies }$ 

${\displaystyle \int \limits _{0}^{\infty }\!t^{n}e^{-t}\,\mathrm {d} t=n\int \limits _{0}^{\infty }\!t^{n-1}e^{-t}\,\mathrm {d} t}$ 

This meets the requirement for the factorial in the sense that:

${\displaystyle n!=n\left(n-1\right)!}$ 

One could then use the logic of Integration by parts and L'Hôpital's rule to conclude that, for integers,

${\displaystyle \int \limits _{0}^{\infty }\!t^{n}e^{-t}\,\mathrm {d} t=n\left(n-1\right)\left(n-2\right)\cdots \left(2\right)\left(1\right)}$ 

One can see that the one in this product correlates to:

${\displaystyle \int \limits _{0}^{\infty }\!e^{-t}\,\mathrm {d} t}$  or

${\displaystyle \int \limits _{0}^{\infty }\!te^{-t}\,\mathrm {d} t,}$ 

depending on how one wants to view it. In this sense, it may even make sense to say that:

${\displaystyle n!=n\left(n-1\right)\left(n-2\right)\cdots \left(2\right)\left(1\right)\left(1\right)}$ 

for any integer.

While this definition of factorial (the integral definition) fits all of the properties of the factorial for integers, one can see that plugging in non-integer values could also make sense. One could examine the case of 1.5 factorial:

${\displaystyle \left(1.5\right)!=\left({\frac {3}{2}}\right)!=\int \limits _{0}^{\infty }\!t^{\frac {3}{2}}e^{-t}\,\mathrm {d} t}$ 

One can apply Integration by parts.

${\displaystyle \int \limits _{C}\!u\,\mathrm {d} v=\left.uv\right|_{C}-\int \limits _{C}\!v\,\mathrm {d} u}$ 

${\displaystyle u=t^{3/2}}$ 

${\displaystyle \mathrm {d} v=e^{-t}\implies v=-e^{-t}}$ 

One can then apply Integration by parts:

${\displaystyle \int \limits _{0}^{\infty }\!t^{\frac {3}{2}}e^{-t}\,\mathrm {d} t=\left.-t^{\frac {3}{2}}e^{-t}\right|_{0}^{\infty }-\int \limits _{0}^{\infty }\!-e^{-t}\left({\frac {3}{2}}t^{\frac {1}{2}}\right)\,\mathrm {d} t=\left.-t^{\frac {3}{2}}e^{-t}\right|_{0}^{\infty }+{\frac {3}{2}}\int \limits _{0}^{\infty }\!t^{\frac {1}{2}}e^{-t}\,\mathrm {d} t}$ 

Using L'Hôpital's rule:

${\displaystyle \left.-t^{n}e^{-t}\right|_{0}^{\infty }=0\implies }$ 

${\displaystyle \left.-t^{\frac {3}{2}}e^{-t}\right|_{0}^{\infty }=0\implies }$ 

${\displaystyle \int \limits _{0}^{\infty }\!t^{\frac {3}{2}}e^{-t}\,\mathrm {d} t={\frac {3}{2}}\int \limits _{0}^{\infty }\!t^{\frac {1}{2}}e^{-t}\,\mathrm {d} t}$ 

This shows that:

${\displaystyle \left({\frac {3}{2}}\right)!={\frac {3}{2}}\cdot \left({\frac {1}{2}}\right)!}$ 

In general, one could use Integration by parts and L'Hôpital's rule to show that:

${\displaystyle \left({\frac {k}{2}}\right)!={\frac {k}{2}}\cdot \left({\frac {k}{2}}-1\right)\cdot \left({\frac {k}{2}}-2\right)\cdots {\frac {3}{2}}\left({\frac {1}{2}}\right)!}$ 

for any odd integer k.

By the extended definition:

${\displaystyle \left({\frac {1}{2}}\right)!=\int \limits _{0}^{\infty }\!t^{\frac {1}{2}}e^{-t}\,\mathrm {d} t}$ 

Using Integration by parts and L'Hôpital's rule implies that:

${\displaystyle \left({\frac {1}{2}}\right)!=\int \limits _{0}^{\infty }\!{\frac {1}{2}}t^{-{\frac {1}{2}}}e^{-t}\,\mathrm {d} t}$ 

One can apply Integration by substitution:

${\displaystyle u=t^{\frac {1}{2}}}$ 

${\displaystyle \mathrm {d} u={\frac {1}{2}}t^{-{\frac {1}{2}}}\mathrm {d} t}$ 

Applying Integration by substitution yields:

${\displaystyle \int \limits _{0}^{\infty }\!{\frac {1}{2}}t^{-{\frac {1}{2}}}e^{-t}\,\mathrm {d} t=\int \limits _{u\left(0\right)}^{\lim \limits _{p\to \infty }u\left(p\right)}\!e^{-u^{2}}\,\mathrm {d} u=\int \limits _{0}^{\infty }\!e^{-u^{2}}\,\mathrm {d} u={\frac {\sqrt {\pi }}{2}}\implies }$ 

${\displaystyle \left({\frac {1}{2}}\right)!={\frac {\sqrt {\pi }}{2}}}$ 

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