Owen's Math Portal

The Gaussian Integral is given by $$ \int_{-\infty}^{\infty} e^{-x^2} dx $$ This is a beautiful integral with many important applications to probability and statistics, namely normal distributions. Here, we will evaluate this integral using techniques from Calculus 3. First, if we define \(I=\int_{-\infty}^{\infty} e^{-x^2} dx\) we have that $$ I^2=\int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} dy dx= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dy dx $$ From here, we will preform a change into polar coordinates, with \((x,y) \rightarrow (t, \theta)\) and \(x=r\cos\theta\) , \(y=r\sin\theta\) . From this, since \(r^2=x^2+y^2\) , \(r \in (0,\infty)\) and \(\theta \in (0, 2\pi)\) we have that our integral is now $$ I^2=\int_{0}^{\infty} \int_{0}^{2\pi} re^{-r^2} d\theta dr $$ after this change of coordinates. By Fubini’s theorem, we have that $$ I^2=\int_{0}^{\infty} \int_{0}^{2\pi} re^{-r^2} d\theta dr = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} re^{-r^2} dr = 2\pi \int_{0}^{\infty} re^{-r^2} dr $$ Now evaulating this integral of one dimension with the substitution $$ t=r^2 \implies \sqrt{t}=r \implies dt=2rdr \implies rdr=\frac{dt}{2} $$ we have that

For a set of orthonormal vectors \((v_1,v_2,...,v_m)\) in an m-dimensional vector space , and set of constants \(r_1,r_2,...r_m \in \mathbb{R}\) , if we consider the equation \(r_1v_1+r_2v_2+...+r_mv_m=0\) with the objective of demonstrating that the vectors in \((v_1,v_2,...,v_m)\) are linearly independent, then we must show that \(r_1=r_2=...=r_m=0\) . If we take the inner product on both sides of the equation , we have that

$$ <r_1v_1+r_2v_2+…+r_mv_m, v_1> = <0,v_1> $$

Using the properties of inner products, we have that

We will evalute the integral \(\int_0^1 \frac{\ln(x)}{1+x} dx \) using Taylor Series expansion and switching the order of integration and summation.

Firstly, we notice that $$ \frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n $$

by taylor series expansion. This series coverges for \(x \in (0,1)\)

so we can substitute this into the integral:

$$ \int_0^1 \frac{\ln(x)}{1+x} dx = \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx $$

Further, by changing the order of integration and summation, we have $$ \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx $$