## Six particle amplitudes from scattering equations

Six-gluon amplitude:

(1) with

(2) ## Integral via Inversions

Let us compute the following integral

(1) Performing , this integral becomes

(2) Thus we have

(3) ## Fresnel integral

(1)  Consider the following integral

(2) From this we have

(3) which gives

(4) Equation (2) also gives

(5) We give an alternative derivation in the following. Let us start with

(6) Differentiating with respect to (7) where we chnage the variable, .
Therefore we have

(8) Since in the limit (9) we have

(10) Solving the equation

(11) gives (4).

## Integral by Euler’s Formula

We compute the following integral

(1) Performing a partial fraction expansion of the integrand

(2) Thus we have

(3) Let us see another integral

(4) Using Euler’s Formula

(5) we have

(6) ## Gaussian integral

The Gaussian integral is:
\begin{align}
\int_{-\infty}^{\infty}dx e^{-x^2} \,=\, \sqrt{\pi}
\end{align}
A standard way to derive this formula is
\begin{align}
\bigg(\int_{-\infty}^{\infty}dx e^{-x^2}\bigg)^2
\,&=\, \int_{-\infty}^{\infty}dx \int_{-\infty}^{\infty}dy\, e^{-(x^2+y^2)}
\\
\,&=\, \int_0^{2\pi} d\phi \int_{0}^{\infty}dr\,r e^{-r^2}
\\
\,&=\, \int_0^{2\pi} d\phi \int_{0}^{\infty}{dr^2 \over 2}\, e^{-r^2}
\\
\,&=\, \pi
\end{align}
Performing the change of variable $$x=\sqrt{t}$$, this turns into the Euler integral
\begin{align}
\int_{-\infty}^{\infty}dx\, e^{-x^2} \,=\, 2 \int_0^\infty dx\, e^{-x^2}
\,=\, 2\int_0^\infty dt\, \frac{1}{2}\ e^{-t} \ t^{-\frac{1}{2}}
\,=\, \Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi}
\end{align}
where $$\Gamma$$ is the gamma function. From this we can see that the factorial of a half-integer is a rational multiple of $$\sqrt{\pi}$$. More generally, let us consider
\begin{align}
\int_0^{\infty} dx\, e^{-\alpha x^\beta}
\end{align}
Change the varible $$t=\alpha x^\beta$$,
\begin{align}
dx \,=\, {1 \over \alpha^{1\over \beta} \beta}\, t^{{1\over\beta} – 1}\, dt
\end{align}
and thus we have
\begin{align}
\int_0^{\infty} dx\, e^{-\alpha x^\beta}
\,=\, {1 \over \alpha^{1\over \beta} \beta}\,\int_0^{\infty} dt\, t^{{1\over\beta} – 1}\,e^{-t} dt
\,=\, {\Gamma\left(\tfrac{1}{\beta}\right) \over \alpha^{1\over \beta} \beta}
\end{align}

Another type of integrals is
\begin{align}
2\int_0^{\infty} dx\,x^{2n} e^{-x^2}
\,=\, \int_0^{\infty} dt\,t^{n-{1\over 2}} e^{-t}
\,=\, \Gamma(n+\tfrac{1}{2})
\end{align}