Question

Given that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{e}}^{\mathbf{-}\mathbf{a}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{k}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}$ differentiate with respect to to show that${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{a}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{2}\mathbf{k}\mathbf{a}}{{\left(a^2+k^2\right)}^{\mathbf{2}}}$ and differentiate with respect to$\mathit{k}$ to show that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{a}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}{{\left(a^2+k^2\right)}^{\mathbf{2}}}$ .

Short answer

${\int }_0^{\infty }x{e}^{-ax}\sin kxdx=\frac{2ka}{{\left(a^2+k^2\right)}^2}$and${\int }_0^{\infty }x{e}^{-ax}\cos kxdx=\frac{a^2-k^2}{{\left(a^2+k^2\right)}^2}$

Step-by-step solution

Given Information

Given that ${\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{k}{a^2+k^2}$ ${\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{k}{a^2+k^2}$ ...(1)

Formula Used

We know that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}$ ...(2)

Solving the given equation

Differentiate the given function (1) with respect to

Compare equation (1) and equation (2).

${\int }_0^{\infty }{e}^{-ax}\sin kxdx=e^{-a\left(\infty \right)}\sin k\infty .\frac{d}{da}\left(\infty \right)-e^{-a\left(0\right)}\sin k\left(0\right).\frac{d}{da}\left(0\right)+{\int }_0^{\infty }\left[e^{-ax}\left(-x\sin kx\right)\right]dx$

${\int }_0^{\infty }{e}^{-ax}\sin kxdx=-{\int }_0^{\infty }x{e}^{-ax}\sin kxdx$ ...(3)

But we know that

$$\begin{gathered} {\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{d}{da}\left(\frac{k}{a^2+k^2}\right) \\ {\int }_0^{\infty }{e}^{-ax}\sin kxdx=k\left(\frac{-1}{{\left(a^2+k^2\right)}^2}\right)\left(2a\right) \end{gathered}$$

....(4)

${\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{-2ak}{{\left(a^2+k^2\right)}^2}$

Therefore, from equations (3) and (4)

${\int }_0^{\infty }x{e}^{-ax}\sin kxdx=\frac{2ka}{{\left(a^2+k^2\right)}^2}$

Now differentiate the equation (1) with respect to

${\int }_0^{\infty }{e}^{-ax}\sin kxdx=e^{-a\left(\infty \right)}\sin k\infty .\frac{d}{dk}\left(\infty \right)-e^{-a\left(0\right)}\sin k\left(0\right).\frac{d}{dk}\left(0\right)+{\int }_0^{\infty }\left[x{e}^{-ax}\cos kx\right]dx$

${\int }_0^{\infty }{e}^{-ax}\sin kxdx={\int }_0^{\infty }x{e}^{-ax}\cos kxdx$ ...(5)

But$$\begin{gathered} \frac{d}{dk}{\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{d}{dk}\left(\frac{k}{a^2+k^2}\right) \\ \frac{d}{dk}{\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{\left(a^2+k^2\right)\left(1\right)-k\left(0+2k\right)}{{\left(a^2+k^2\right)}^2} \end{gathered}$$

$\frac{d}{dk}{\int }_0^{\infty }{e}^{-ax}\sin kxdx=\frac{\left(a^2-k^2\right)}{{\left(a^2+k^2\right)}^2}$ ...(6)

Therefore,${\int }_0^{\infty }x{e}^{-ax}\cos kxdx=\frac{a^2-k^2}{{\left(a^2+k^2\right)}^2}$