91Ó°ÊÓ

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

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

Differentiate the given function (1) with respect to

Compare equation (1) and equation (2).

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

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

But we know that

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

....(4)

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

Therefore, from equations (3) and (4)

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

Now differentiate the equation (1) with respect to

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

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

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

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

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