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Question Number 128485 by BHOOPENDRA last updated on 07/Jan/21

$$f\left(t\right)=\frac{1-cos2t}{t}$$$$findlaplacetransformation?$$

Answered by Dwaipayan Shikari last updated on 07/Jan/21

$$\mathcal{L}\left(\frac{1-cos2t}{t}\right)\Rightarrow I\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{1-cos2t}{t}dt$$$$I^{\prime }\left(s\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-st}dt+{\int }_0^{\mathrm{\infty }}{e}^{-st}cos2t=-\frac{1}{s}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-\mathit{s}\mathit{t}+2\mathit{i}\mathit{t}}+{\mathit{e}}^{-\mathit{s}\mathit{t}-2\mathit{i}\mathit{t}}$$$$=-\frac{1}{\mathit{s}}+\frac{1}{2}(\frac{1}{\mathit{s}-2\mathit{i}}+\frac{1}{\mathit{s}+2\mathit{i}})=-\frac{1}{\mathit{s}}+\frac{\mathit{s}}{{\mathit{s}}^2+4}$$$$\mathit{I}\left(s\right)=\mathit{l}\mathit{o}\mathit{g}\left(\frac{\sqrt{{\mathit{s}}^2+4}}{\mathit{s}}\right)+\mathit{C}\Rightarrow \mathit{I}\left(\mathit{s}\right)=\mathit{l}\mathit{o}\mathit{g}\left(\frac{\sqrt{{\mathit{s}}^2+4}}{\mathit{s}}\right)(\mathit{C}=0)$$

Commented by BHOOPENDRA last updated on 07/Jan/21

$$thankssir$$

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