While proving that $$\int^{\infty}_0 \frac{\sin x}xdx$$
I saw the Laplace Transform proof.
It used that $$\cal L\left\{\frac{\sin t}{t}\right\}=\int^\infty_0 \cal L\left\{\sin(t)\right\}d\sigma$$
So for understanding it, I tried:
$$\cal L\left\{\frac{\sin t}{t}\right\}=\int^\infty_0e^{-st}\frac{\sin t}{t}dt=\int^\infty_0\frac1t\cal L\left\{\sin t\right\}dt$$
But I cannot see how that $\sigma$ emerged and $t^{-1}$ vanished?
Also, how do we know that using the Laplace transform, we would get an integral that is equal to the original one ($\int^\infty_0\frac{\sin x}{x}dx$)
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asked Dec 30, 2015 at 9:53
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2 Answers
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You can write,
$$ \int_{0}^{\infty}\frac{\sin t}{t}dt=L\bigg\{\frac{\sin t}{t}\bigg\}_{s=0}=\Bigg(\int_{s}^{\infty}L\big\{\sin t\big\}du\Bigg)_{s=0}\\ =\Bigg(\int_{s}^{\infty}\frac{1}{1+u^{2}}du\Bigg)_{s=0}=\Bigg(\tan^{-1}(u)\Bigg|^{\infty}_{s}\Bigg)_{s=0}=\frac{\pi}{2}. $$
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$\begingroup$ How did you deduce the third expression? $\endgroup$ Commented Dec 30, 2015 at 10:39
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$\begingroup$ if $L\{f(t)\}=F(s)$ and $f(t)/t\rightarrow \alpha<\infty$ as $t\rightarrow 0^{+}$ , then $$L\{f(t)/t\}=\int_{s}^{\infty}F(u)du$$. This is one of the well known properties of the Laplace transform. $\endgroup$– AlbertCommented Dec 30, 2015 at 10:42
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Your formula is a (too) short notation, suppressing the variable of the Laplace transform. It should read $$\mathcal{L}\left\{\frac{\sin t}{t} \right\} (0) = \int_0^\infty \mathcal{L}\{ \sin t\}(\sigma) \,d\sigma.$$
This follows from the rule `Frequency-domain integration'. A proof of this is rather straightforward. If you have troubles, I can provide some more help.
