How to prove ∫0∞e−x2dx=π2

Question

How to prove  ∫0∞e−x2dx=π2

movingsupplyw1 · Accepted Answer

This is an old favorite of mine.
Define

I = \int_{- \infty}^{+ \infty} e^{- x^{2}} d x

Then

I^{2} = (\int_{- \infty}^{+ \infty} e^{- x^{2}} d x) (\int_{- \infty}^{+ \infty} e^{- y^{2}} d y)

I^{2} = \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} e^{- (x^{2} + y^{2})} d x d y

Now change to polar coordinates

I^{2} = \int_{0}^{+ 2 π} \int_{0}^{+ \infty} e^{- r^{2}} r d r d θ

The

θ

integral just gives

2 π

, while the r integral succumbs to the substitution

u = r^{2}

I^{2} = 2 π \int_{0}^{+ \infty} e^{- u} d \frac{u}{2} = π

So

I = \sqrt{π}

and your integral is half this by symmetry
I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

Carl Swisher · Accepted Answer

We can start again with the observation

(\int_{- \infty}^{\infty} e^{- x^{2}} d x) (\int_{- \infty}^{\infty} e^{- y^{2}} d y) = (\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- (x^{2} + y^{2})} d x d y) = V

Now V is simply the volume of the body

- \infty < x, y < \infty

0 < z < e^{- (x^{2} + y^{2})}

or, equivalently,

0 < x^{2} + y^{2} < - \ln z

0 < z < 1

This implies that the body is a solid of revolution. Using the disk integration formula, we have

V = \int_{0}^{1} π (- \ln z) d z = {[π (z - z \ln z)]}_{0}^{1} = π

Vasquez · Accepted Answer

Define f and g as:f(x):=(∫0xe−t2dt)2 and g(x):=(∫01e−x2(t2+1)t2+1dt)Now,f′(x)=2e−x2∫0xe−t2dtandg′(x)=∫01ddx[e−x2(t2+1)t2+1]dt=−2xe−x2∫01e−x2t2dtSo putting t=tx get ∫01e−x2t2dt=1x∫0xe−t2dtThen we get:g′(x)=−2e−x2∫0xe−t2Thus f&#039;(x)+g&#039;(x)=0 for all x, then f(x)+g(x) is an constant function. Alsof(0)+g(0)=∫011t2+1dt=π4Then f(x)+g(x)=π4 for all x.NSKNow limx→+∞g(x)=0Soπ4=limx→+∞f(x)+g(x)=limx→+∞f(x)=(∫0∞e−t2dt)2Thus∫0∞e−t2dt=π4=π2

How to prove \int_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}

Answered question

Answer & Explanation

New Questions in Calculus 1