Question

Given that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{-}\frac{\mathbf{d}\mathbf{x}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}\mathbf{y}}$, differentiate with respect to y and so evaluate${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{-}\frac{\mathbf{dx}}{{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{.}$

Short answer

The value of ${\int }_0^{\infty }-\frac{\mathrm{dx}}{{\mathrm{y}}^2+{\mathrm{x}}^2}$is $\frac{\mathrm{\pi }}{4y^3}$.

Step-by-step solution

Given Information

Given that${\int }_0^{\infty }-\frac{\mathrm{dx}}{{\mathrm{y}}^2+{\mathrm{x}}^2}=\frac{\mathrm{\pi }}{2\mathrm{y}}$.

Formula Used

We know that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}$.

Evaluate ∫0∞-dx(y2+x2).

$$\begin{gathered} {\int }_0^{\infty }\frac{dx}{y^2+x^2}dx=\frac{1}{y^2+{\infty }^2}.\frac{d}{dy}\left(\infty \right)-\frac{1}{y^2+0^2}.\frac{d}{dy}\left(0\right)+\int \frac{-1}{\left(y^2+x^2\right)}\left(2y\right)dx \\ =-2y{\int }_0^{\infty }\frac{dx}{\left(y^2+x^2\right)} \end{gathered}$$

But,

$$\begin{gathered} {\int }_0^{\infty }\frac{dx}{y^2+x^2}dx=\frac{\mathrm{\pi }}{2y} \\ {\int }_0^{\infty }\frac{dx}{y^2+x^2}dx=\frac{d}{dy}\left(\frac{\mathrm{\pi }}{2y}\right) \\ =-\frac{\mathrm{\pi }}{2y^2} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{-\mathrm{\pi }}{2y^2}-=-2y{\int }_0^{\infty }\frac{dx}{{\left(y^2+x^2\right)}^2} \\ {\int }_0^{\infty }\frac{dx}{{\left(y^2+x^2\right)}^2}=\frac{-\mathrm{\pi }}{\left(2y^2\right)\left(-2y^2\right)} \\ {\int }_0^{\infty }\frac{dx}{{\left(y^2+x^2\right)}^2}=\frac{\mathrm{\pi }}{4y^3} \end{gathered}$$

Hence ${\int }_0^{\infty }\frac{dx}{{\left(y^2+x^2\right)}^2}=\frac{\mathrm{\pi }}{4y^3}$.