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Evaluate $\int_0^{\infty{}}e^{-2t}\sinh{t\frac{\sin t}{t}}\ dt$
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(\begin{align} \displaystyle Consider,\[2ex] L\left[\frac{\sin{t\ }}{t}\right]&=\int_s^{\infty{}}\frac{1}{1+s^2}ds \[2ex] \displaystyle &={\left[\tan^{-1}s\right]}_s^{\infty{}}\[2ex] &=\tan^{-1}{\infty{}-\tan^{-1}s}\[2ex] &= \frac{\pi{}}{2}-\tan^{-1}s\[2ex] &=\cot^{-1}{s} \[2ex] \end{align}\[2ex] )

\displaystyle \sinh t={\ \frac{{(\ e}^t-e^{-t})}{2}} \\[2ex] \begin{align*} \displaystyle \therefore{}\ L \ \left[\frac{{(\ e}^t-e^{-t})}{2}\frac{\sin{t\ }}{t}\ \right]& = \ \frac{1}{2} \left(L\left[e^t\frac{\sin{t\ }}{t}\right], \ L\left[e^{-t}\frac{\sin{t\ }}{t}\right]\right) \\[2ex] \displaystyle &= \ \frac{1}{2}\left(\cot^{-1}{\left(s-1\right)}-\cot^{-1}{\left(s+1\right)}\right) \\[2ex] \end{align*}\\ $\displaystyle {L\left[e^{ax}f\left(x\right)\right]=\ F(s-a)} \[2ex] \displaystyle \ \therefore{}L\ \left[\sin ht\frac{\sin t}{t}\right]=\frac{1}{2}\left(\cot^{-1}{\left(s-1\right)}-\cot^{-1}{\left(s+1\right)}\right)…\ …\ (A) \[3ex]From \ the \ definition \ of \ Laplace \ Transform,\[3ex] \$ $\displaystyle L\left[f\left(x\right)\right]=\int_0^{\infty{}}\ e^{-sx}f\left(x\right)dx \[2ex] \displaystyle \therefore{}\ L\ \left[\sin ht\frac{\sin t}{t}\right]=\ \int_0^{\infty{}}\ e^{-st}\sin ht\frac{\sin t}{t}d \[2ex] \displaystyle \therefore{}\ \int_0^{\infty{}}\ e^{-st}\sin ht\frac{\sin t}{t}dt=\ \frac{1}{2}\left(\cot^{-1}{\left(s-1\right)}-\cot^{-1}{\left(s+1\right)}\right) \[4ex] {From (A)} \ Put\ s=2 }\[4ex]$

(\begin{align} \displaystyle \therefore{}\ \int_0^{\infty{}}\ e^{-2t}\sin ht\frac{\sin t}{t}dt&= \ \frac{1}{2}\left(\cot^{-1}{\left(2-1\right)}-\cot^{-1}{\left(2+1\right)}\right) \[2ex] \displaystyle &= \ \frac{1}{2}\left(\cot^{-1}1-\cot^{-1}3\right) \[2ex] \displaystyle &= \ \frac{1}{2}\left(\frac{\pi{}}{4}-\cot^{-1}3\right) \[2ex] \end{align})

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