Consider a point P, a unit away from the long charged wire. Electric field due to element dy, $dE_p$ = $\frac{Q}{4\pi\varepsilon_0 r^2}$ or Q = λdy $r^2$ = $y^2+a^2$ ⇒ $dE_p$ = $\frac{\lambda dy}{4\pi\varepsilon_0(y^2+a^2)}$ ⇒ $E_p$ = $\int\limits_{-\infty}^{\infty}$ $dE_p$ Vertical components cancel out and horizontal components are added due to symmetry ⇒ $E_p$ = $\int\limits_{0}^{\infty}$ $\frac{2\lambda dy}{4\pi\varepsilon_0(y^2+a^2)}$ × $(\cos\theta)_1$ = $\frac{2}{4\pi\varepsilon_0} \int\limits_{0}^{\infty} \frac{\lambda dy}{y^2+a^2}$ × $\frac{y}{\sqrt{y^2+a^2}}$ = $\frac{2}{4\pi\varepsilon_0} \int\limits_{0}^{\infty} \frac{\lambda y dy}{(y^2+a^2)^{3/2}}$ Take $y^2+a^2$ = x ⇒ 2ydy = dx Taking proper limits $E_p$ = $\frac{\lambda}{4\pi\varepsilon_0} \int\limits_{a^2}^{\infty} \frac{dx}{x^{3/2}}$ ⇒ $\left[ \frac{-2\lambda}{4\pi\varepsilon_0} \frac{1}{\sqrt{x}} \right]_{a^2}^{\infty}$ = $λ/{2πε_0a}$