# problem

Hello, this is sudden, but a problem.

i! \ overline {i! } = 0.27 Show

This is not a riddle. These are the mathematical formulas listed. Even those who are taking special function theory may not even think of it as a mathematical formula for a moment. The root of all evil is the font. The problem statement can be rewritten with the $\ TeX$ characters as follows.

Show $i! \ Overline {i!} = 0.27$

In this case, you can clearly understand the problem statement. The reason for adding $\ overline {}$ is to make the calculation meaningful. It looks like a scam, but please refrain from it.

The point is to find the factorial of the imaginary number.

We are still at the stage where we can decipher the problem statement. First of all, I'm riding on an imaginary floor! That's it. Calculating only with integers on the floor Normally, it is calculated. For integers $1,1,2,6,24,\cdots$

It is a sequence that follows. When written in recurrence formula,

a_n=n a_{n-1}

It becomes. Because it is difficult to suddenly extend the concept of factorial to complex numbers Extends to real numbers.

# Gamma function

A method whose input is a single argument and an integer is called a sequence. A method whose input is a single argument and is a real number is called a function.

If you extend the domain to a real number, it becomes a function (note that it may not be the case).

Find a smooth curve that complements $a_j$ and $a_ {j + 1}$ in the sequence. I happened to find the gamma function.

The gamma function $\ Gamma (x)$ is known to have properties very similar to factorial. The gamma function $\ Gamma (x)$ can be said to be a generalization of factorial.

I don't understand the meaning even if I look at the definition formula of the gamma function $\ Gamma (x)$. Let's look at some properties

1. $\ Gamma (n) = (n-1)!$ For a positive integer $n$
2. $\ Gamma (x + 1) = x \ Gamma (x)$ for the positive real number $x$
3. \Gamma(1)=1

Looking at these, you can be convinced that the gamma function is a generalization of factorial. The second fact is the factorial recurrence formula. First, be aware that $n$ is a little off The third only defines the first term. In fact, the input of the gamma function can be calculated well not only with real numbers but also with complex numbers whose real parts are positive. Now you can think of $i!$.

If you go a little deeper into the gamma function and study, you will come across a conflicting formula. Gamma function reciprocity formula

\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin \pi x}

The official proof of the conflict formula is difficult, but I think that those who have studied the residue theorem will enjoy reading the proof.

Please note that there are some amakudari parts. First, set $z = ix$. The reason why $x$ can be attached is that if you make a general formula and substitute it at the end, the calculation will be successful. (Same for Euler's formula) $|z!|^2=|\Gamma(z+1)\Gamma(z+1)|=|z||\Gamma(z)\Gamma(z+1)|.$ here, $|\Gamma(z)\Gamma(z+1)|=\Gamma(z)\Gamma(1-z)$ Is used. (This proof is simple and boring, so I'll omit it.) Applying to the formula below, $|z!|^2=z\Gamma(z)\Gamma(1-z)$ Also, from the conflict formula $|z!|^2=\cfrac{\pi z}{\sin \pi z}$ Substituting $z = ix$ $(ix)!\overline{(ix)!}=(ix)!^2=\cfrac{i\pi x}{\sin i\pi x}$
It becomes. Using $\ sin ix = i \ sinh x$ and substituting x = 1 $i!\overline{i!}=(ix)!^2=\cfrac{\pi x}{\sinh \pi x}\mid_{x=1}$ If you leave the rest to the calculator $\cfrac{\pi}{\sinh \pi}=0.27$ And $i!\overline{i!}=0.27$
I was able to show. If you use the gamma function technically, $\int_0^1 x^x dx$ You can also calculate something.