Electrostatics: Electric field
Some integrals

1. Integral: ∫dx/(x² + R²)^3/2

Let's find the following integral:
$$\Large\bf\color{brown}{ I = \int\limits_{-\infty}^{+\infty} \frac{dx}{(x^2 + R^2)^{\frac{3}{2}}} }$$ Let the following change of variable:

x/R = tg y

When x goes from - ∞ towards + ∞,
y goes from - π/2 towards + π/2.

Taking the derivative:
dx = R tg'y dy = R (1 + tg²y) dy

x² + R² = R² tg²y + R² = R²(1 + tg²y)

∫dx/(x² + R²)^3/2 = ∫ R (1 + tg²y) dy/(x² + R²)^3/2
= ∫ R (1 + tg²y) dy/ R³(1 + tg²y)^3/2
= ∫ dy/ R²(1 + tg²y)^1/2 = ∫ dy cos y/ R²

This integral is even, It follows that:

∫ dx/(x² + R²)^3/2 = 2 (1/ R²) ∫ dy cos y
= 2 (1/ R²) sin y (y: from 0 to + π/2) = 2/R²

$$\Large\bf\color{green}{ \int\limits_{-\infty}^{+\infty} \frac{dx}{(x^2 + R^2)^{\frac{3}{2}}} = \frac{2}{R^2} }$$

Now let's find the following integral:

$$\Large\bf\color{brown}{ I = \int\limits_{-l}^{+l} \frac{dz}{(z^2 + y^2)^{\frac{3}{2}}} }$$
Let the following change of variable:

z/y = tg θ

When z goes from - l towards + l,
θ goes from - arctan(l/y) towards + arctan(l/y).

from above, we have:

I = 2 (1/ y²) sin θ
from 0 towards + arctan(l/y)

We have

sin²θ = 1 - cos²θ = 1 - sin²θ/tan²θ, then
sin θ = tan θ/√(1 + tan²θ)

Hence

I = 2 (1/ y²) sin θ
θ : from 0 towards + arctan(l/y)

= 2 (1/ y²) [(sin(arctan(l/y) - sin(0)] =
2 (1/ y²) [(l/y)/√(1 + (l/y)²)] = (2l/ y²)/√(y² + l²)

$$\Large\bf\color{blue}{ \int\limits_{-l}^{+l} \frac{dz}{(z^2 + y^2)^{\frac{3}{2}}} = \frac{2l}{y^2 \sqrt {y^2 + l^2}} }$$

2. Integral: ∫exp(ax) cosbx dx

We consider the following integral:

∫exp(ax) cosbx dx

I = ∫exp(ax) cos bx dx

By part;

exp(ax) = u and
dv = cos bx → v = (1/b)sin bx

I = exp(ax) (1/b)sin bx - ∫ (a/b)sin bx exp(ax) dx
= exp(ax) (1/b)sin bx - (a/b) ∫ sin bx exp(ax) dx

J = ∫ sin bx exp(ax) dx
u = exp(ax) dv = sin bx dx → v = - (1/b) cos bx
J = - (1/b) cos bx exp(ax) + (a/b)∫ exp(ax) cos bx
= - (1/b) cos bx exp(ax) + (a/b)I

I = exp(ax) (1/b)sin bx - (a/b)J
= exp(ax) (1/b)sin bx - (a/b)[- (1/b) cos bx exp(ax) + (a/b)I]
= exp(ax) (1/b)sin bx + (a/b)[ (1/b) cos bx exp(ax) - (a/b)I]
= exp(ax) (1/b)sin bx + (a/b²) cos bx exp(ax) - (a²/b²)I

(a² + b²)/b² I = exp(ax) (1/b)sin bx + (a/b²) cos bx exp(ax) (a² + b²) I
= exp(ax) b sin bx + a cos bx exp(ax)

I = [b sin bx exp(ax) + a cos bx exp(ax)]/(a² + b²)

3. The probability Integrale

J(n) = ∫ xⁿ exp(- a x²)dx     (1)
[- ∞ → + ∞]

The function to integrate : xⁿ exp(- a x²) is odd and therefore the integrale J(n) is null if n is odd.

It is even if n is even and therefore the integrale J(n) can be written as:

J(n) = 2. I(n)     (2)

Where:

I(n) = ∫ xⁿ exp(- a x²) dx [0 → + ∞]

I(n) = ∫ xⁿ exp(- a x²)dx     (3)
[0 → + ∞]

First, Let's integrate I(0).

I(0) = ∫ exp(- a x²) dx [0 → + ∞]
x is a dummy variable, thus:

I(0) = ∫ exp(- a y²) dy [0 → + ∞]

Multiplying the two equations, we get;

I²(0) = ∫ exp(- a x²) . ∫ exp(- a y²) dy [0 → + ∞]
= ∫ ∫ exp(- a x² . exp(- a y² dx dy [0 → + ∞] =
∫ ∫ exp[- a (x² + y²)] dx dy [0 → + ∞]     (4)


Let's switch to the polar coordinates:

r² = x² + y²
dxdy = r dr dθ



The equation (4) becomes:

I²(0) = ∫ exp[- a r²] r dr dθ [0 → + ∞ & 0 → π/2]     (5)

(remark that the bouderies of the integrale are 0 and π/2 because r is limited to the first quadrant)

The equation (5) becomes:

I²(0) = (π/2) ∫ exp[- a r²] r dr [0 → + ∞]     (6)

∫ exp[- a r²] r dr [0 → + ∞] is straightforward.
Let's substitute this way:

a r² = z. Then dz = 2 a r dr. and r dr = dz/2a

Then: ∫ exp[- a r²] r dr = (1/2a)∫ exp[- z] dz =
- (1/2a)∫ exp[- z] [0 → + ∞] = 1/2a

The equation (6) becomes:

I²(0) = (π/2) (1/2a)= π/4a     (7)

And:

I(0) = (π/2) (1/2a)= [π/4a]^1/2     (8)


I(1) can be written as :

I(1) = ∫ x exp(- a x² dx [0 → + ∞]     (9)

From the relationships (6) and (7), it comes:

I(1) = (2/π)I²(0) = (2/π)I²(0) = (2/π)π/4a = 1/2a

I(1) = 1/2a     (10)


Now we will find I(2).

I(2) = ∫ x² exp(- a x² dx [0 → + ∞]     (11)

Let's take the derivative of I(0):

dI(0)/da = d[∫ exp(- a x² dx]/da [0 → + ∞] =

- ∫ x² exp(- a x² dx] [0 → + ∞] = - I(2)     (12)

From the equation (8), we get:

dI(0)/da = d([π/4a]^1/2)/da = (1/2)(π/4)^1/2)(a)^1/2 (-1/a²) =
- (π)^1/2/4 a^3/2     (13)

Equating (12) and (13), we get:

- I(2) = - (π)^1/2/4 a^3/2

Then:

I(2) = (π)^1/2/4 a^3/2     (14)


The derivative of I(1)is:

dI(1)/da = d[∫ x exp(- a x² dx]/da [0 → + ∞] =
- ∫ x³ exp(- a x² dx = - I(3)

From the equation (10), we get:

dI(1)/da = -1/2a². Equating, we get:

- I(3) = -1/2a². Then:

I(3) = 1/2a²     (15)

We can continue this way : taking the derivative of
I(n) to get - I(n + 2) and equating with dI(n)/da in
order to find I(n + 2).

Resuming, we have :

I(0) = [π/4a]^1/2
I(1) = 1/2a
I(2) = (π)^1/2/4 a^3/2
I(3) = 1/2a²
...
By recurrence, the following is inferred:

For odd n:
I(n) = [(n - 1)/2]!/ 2 a^{(n + 1)/2}

For even n:
I(n) = [π/a]^1/2 [1.3.5. ... .(n - 1)]/2.(2a)^n/2