Gaussian Integral DRAFT


I=e12x2dxJ=0ex2dxK=eπx2dx\begin{aligned} I & = \int^{\infty}_{-\infty} e^{-\frac 1 2 x^2} dx\\ J & = \int^{\infty}_{0} e^{-x^2} dx\\ K & = \int^{-\infty}_{\infty} e^{-\pi x^2}dx \end{aligned}

Theorem: I=2πI = \sqrt{2\pi}, and, equivalently, J=π/2J = \sqrt{\pi}/2 and K=1K = 1.

Proof by Change of Variables

Alternative Proof by Change of Variables

According to the first proof, we have

J2=0(0e(x2+y2)dx)dyJ^2 = \int^{\infty}_{0} (\int^{\infty}_{0} e^{-(x^2+y^2)} dx)dy\\

Set x=ytx = yt

J2=0(0ey2(1+t2)y dt)dy=0(0y ey2(1+t2) dy)dt \begin{aligned} J^2 & = \int^{\infty}_{0} (\int^{\infty}_{0} e^{-y^2(1+t^2)} y\ dt)dy\\ & = \int^{\infty}_{0} (\int^{\infty}_{0} y\ e^{-y^2(1+t^2)}\ dy)dt\ \end{aligned}

Interchange of integral justifies, by Fubini's theorem for improper Riemann integrals, integration by parts brah brah brah

0y eay2 dy=12a\int^{\infty}_{0} y\ e^{-ay^2}\ dy = \frac 1 {2a} J2=0dt2(t2+1)=12π2=π4J^2 = \int^{\infty}_{0} \frac {dt}{2(t^2+1)} = \frac 1 2 \cdot \frac \pi 2 = \frac \pi 4

Proof by Differentiating under Integral Sign

Proof by Volume Integral

Proof by Gamma Function

n!=0tnetdtn! = \int^{\infty}_{0} t^n e^{-t}{\mathrm d}t Gamma function

Γ(x)=0txetdtt,\Gamma (x) = \int^{\infty}_{0} t^x e^{-t} \frac {\mathrm d t} t, Γ(x)Γ(y)Γ(x+y)=01tx1(1t)y1dt\frac {\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int^{1}_{0} t^{x-1}(1-t)^{y-1}\mathrm d t

Proof by Asymptotic Estimate

Laplace's Original Proof

Proof by Residue Theorem

◀ Divisibility and Greatest Common DivisorsThe Fast Fourier Transform Intuition ▶