Half integral of x

In my previous post I showed you how to take the half derivative of x. Here, I am going to explain how to find the half integral of x.

We can write the half integral of x as

$\int_\frac{1}{2}f(x)\,(dx)^\frac{1}{2}$

This is a way I’ve come up with to represent the half integral of a function, as I couldn’t find anything about it.

We can write the second, third, fourth integral of a function as

$\int\int&space;f(x)\,(dx)^2$ $\int\int\int&space;f(x)\,(dx)^3$ $\int\int\int\int&space;f(x)\,(dx)^4$

and, in general,

$\int\int...\int&space;f(x)\,dx\,dx...\,dx$

or

$\int_n&space;f(x)\,(dx)^n$

So we will write the half integral of x like this:

$\int_\frac{1}{2}&space;x\,(dx)^\frac{1}{2}$

Here we found that

$\frac{\mathrm{d}&space;^n}{\mathrm{d}&space;x^n}\,x^{2n}=\frac{2(2n-1)!}{(n-1)!}\,x^n$

And, in particular, that

$\frac{\mathrm{d}&space;^\frac{1}{2}}{\mathrm{d}&space;x^\frac{1}{2}}\,x=\frac{2}{\sqrt{\pi}}\sqrt&space;x$

This is the half derivative of x, i.e. the inverse of the half integral of x.

We know that the integral of x is x²/2 and that the derivative of x² is 2x. As you can see, the coefficient of the x in the result of the integral is the reciprocal of that of the derivative. This means that the coefficient of the x in result of the half integral of x is

$\left&space;(\frac{2(2n-1)!}{(n-1)!}&space;\right&space;)^{-1}=\frac{(n-1)!}{2(2n-1)!}$

What we need to find now is the exponent of the x. We can give a look at these integrals:

$\int&space;x&space;\,dx=\frac{1}{2}x^2+C$ $\int\int&space;x&space;\,(dx)^2=\frac{1}{6}x^3+C$

As you can see, the exponent of the x in the result is n (the order of the integral) + 1. This means the coefficient of the x for the half integral is 1/2+1=3/2.

Overall,

$\int_n&space;x^{2n}\,(dx)^n=\frac{(n-1)!}{2(2n-1)!}\,x^{n+1}&space;+&space;C$

Therefore,

$\int_\frac{1}{2}x\,(dx)^\frac{1}{2}=\frac{\left&space;(&space;-\frac{1}{2}&space;\right&space;)!}{2}x^\frac{3}{2}+C$

In the previous post we saw that (-0.5)! = sqrt(pi), so

$\int_\frac{1}{2}x\,(dx)^\frac{1}{2}=\frac{\sqrt\pi}{2}x^\frac{3}{2}=\frac{\sqrt\pi}{2}x\sqrt&space;x+C$

I really hope you liked this lesson and, if you have any questions, leave a comment and I will be happy to help!