Focused Electric Field

Recall that $dE_y = \dfrac{k \, dq}{r^2} \cos \theta$ .

So, what's left to do is to prove that $\displaystyle \int_0^\infty \dfrac{\sqrt{1+4x^2} \left(\frac14 - x^2\right) }{ \left[ x^2 + \left(\frac14 - x^2\right)^2 \right] ^{3/2} } \, dx = \dfrac83$ .

Note that we can simplify the denominator of the integrand to

$\left[ x^2 + \left(\dfrac14 - x^2\right)^2 \right ] ^{3/2} = \left[x^2 + x^4 + \dfrac1{16} - \dfrac12 x^2 \right]^{3/2} = \left[ x^4 + \dfrac1{16} + \dfrac12 x^2 \right]^{3/2} = \left[ \left(x^2 + \dfrac14\right)^2 \right]^{3/2} = \left (x^2 + \dfrac14\right)^3$

We start by making the trigonometric substitution $x = \dfrac12\tan y$ , then $\dfrac{dx}{dy} = \dfrac12 \sec^2 y$ ,

$\begin{aligned} \int_0^\infty \dfrac{\sqrt{1+4x^2} \left(\frac14 - x^2\right) }{ \left[ x^2 + \left(\frac14 - x^2\right)^2 \right] ^{3/2} } \, dx &=& \int_0^\infty \dfrac{\sqrt{1+4x^2} \left(\frac14 - x^2\right) }{ \left( x^2 + \frac14\right)^3 } \, dx \\ &=& \int_0^{\pi /2} \dfrac{\sqrt{1+\tan^2 y} \cdot \frac14 (1 - \tan^2 y)}{\left(\frac14\right)^3 (1+ \tan^2 y)^3 } \cdot \dfrac12 \sec^2 y \, dy \\ &=& 8 \int_0^{\pi /2} \dfrac{\sec y (1-\tan^2 y)}{(\sec^2 y)^3} \cdot \sec^2 y \, dy \\ &=& 8 \int_0^{\pi /2} \dfrac{1- \tan^2 y}{\sec^3 y } \, dy \\ &=& 8 \int_0^{\pi /2} \cos y (1 - 2\sin^2 y ) \, dy \qquad, \qquad \text{ let } y = \sin y \Rightarrow dy = \dfrac{dz}{\cos z} \\ &=& 8 \int_0^1 (1 - 2z^2) \, dz \\ &=& 8 \left [ z - \dfrac23 z^3 \right]_0^1 = \dfrac83 \; . \end{aligned}$

And we're done!