Proof for Area under Gaussian distribution

This section concerns itself with finding the integral of a Gaussian Distribution

∫−∞∞e−x2 dx=C

Find C.

Proof

Lets consider adding another dimension.

f(x,y)=e−x2−y2=e−x2⋅e−y2=e−r2

where r2=x2+y2. Let r be the radius of the circle. Thus all points a distance r from the origin on e−x2−y2 would have the same value. We can graph out the function against x and y
Pasted image 20250506202742.png|600

We can find the volume under f(x,y) by summing the area under the red line (points of equidistant from the origin) for all values r. Unroll the area under the red line to form a rectangle, with length 2πr and height e−r2, integrate with respect to r.

V=∫0∞2πr e−r2 dr=π∫0∞2r e−r2 dr=π[−e−r2 ]0∞=π

We can express the volume as a double integral as below, where the inner integral integrates in the x axis, keeping y as a constant. I like to think of the integral as infinitely slicing f(x,y) along the x axis and finding the area bounded by f(x,y) in each slice, expressing it as some value. Next the outer integral sums up all these areas by integrating in the y direction, thus obtaining the volume (V) under f(x,y).

V=∫−∞∞∫−∞∞e−x2−y2 dx dy=∫−∞∞∫−∞∞e−x2e−y2 dx dy=∫−∞∞e−y2∫−∞∞e−x2 dx dy[We can factor e−y2 as it is a constant]=∫−∞∞e−y2C dy=C∫−∞∞e−y2 dy=C2

Thus

V=C2=π∫−∞∞e−x2 dx=C=π
Thus

∫−∞∞e−x2 dx=π

Source: https://www.youtube.com/watch?v=cy8r7WSuT1I