1
$\begingroup$

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 ($\int^\infty_0\frac{\sin x}{x}dx$)

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

You can write,

$$ \int_{0}^{\infty}\frac{\sin t}{t}dt=L\bigg\{\frac{\sin t}{t}\bigg\}_{s=0}=\Bigg(\int_{s}^{\infty}L\big\{\sin t\big\}du\Bigg)_{s=0}\\ =\Bigg(\int_{s}^{\infty}\frac{1}{1+u^{2}}du\Bigg)_{s=0}=\Bigg(\tan^{-1}(u)\Bigg|^{\infty}_{s}\Bigg)_{s=0}=\frac{\pi}{2}. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you deduce the third expression? $\endgroup$ – Aditya Agarwal Dec 30 '15 at 10:39
  • $\begingroup$ 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. $\endgroup$ – Albert Dec 30 '15 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.