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I need to find the Laplace transform of $$\frac{\sin(t)}{t}$$ without using the following rule $\mathcal{L}\left(\frac{f(t)}{t}\right)=\int_s^{\infty}F(u)du$.

We aren't allowed to use this rule unless we can prove it, and I'm assuming our lecturer does not want us to use it.

However, we were given the following hint "For this question, you may need to evaluate $ \int^\infty_0 \frac{\sin(t)}{t}dt$. Try substituting $\frac{1}{t} = \int^\infty_0 e^{-st} ds$."

I'm not sure what exactly this is supposed to mean and would appreciate it if someone could show me what this would look like. Does it mean evaluating $ \int^\infty_0 e^{-st} \int^\infty_0 e^{-st}\sin(t) dsdt$? I'm really confused.

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  • $\begingroup$ Related post: math.stackexchange.com/questions/1247365/… (not answered yet). Searched from approach0.xyz/search/… $\endgroup$
    – Wei Zhong
    Sep 10 '16 at 12:56
  • $\begingroup$ I'm pretty sure I don't need to find a series solution for this. Also, there's a hint here that suggests otherwise. But I'm not sure how to interpret the hint. $\endgroup$ Sep 10 '16 at 12:58
  • $\begingroup$ @Moo, the answers on that question also don't do what the hint suggests. $\endgroup$ Sep 10 '16 at 13:00
  • $\begingroup$ Following the hint, $$\int_0^\infty e^{-st}\frac{\sin t}t dt=\int_0^\infty e^{-st}\sin t\int_0^\infty e^{-xt}dxdt=\int_0^\infty \left(\int_0^\infty e^{-(s+x)t}\sin tdt\right) dx$$ Now the Laplace transform of the sine is ... hence the inner integral is ... $\endgroup$
    – Did
    Sep 10 '16 at 13:00
  • $\begingroup$ Ohhhhhh, thank you so much @Did that makes sense! I'd like to give you the correct answer. $\endgroup$ Sep 10 '16 at 16:56
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I think you should write $$ \int_0^\infty e^{-ut} \frac{\sin{t}}{t} = \int_0^\infty \int_0^\infty e^{-(u+s)t} \frac1{2i}( e^{it}-e^{-it}) \;ds \; dt$$ and change the order of integration. Details depends on the expected level of math rigor.

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Assuming $s>0$, the functions $\frac{\sin t}{t}e^{-st}$ and $\sin(t)e^{-st}$ belong to $L^1(\mathbb{R}^+)$ and $$ g(s)=\int_{0}^{+\infty}\frac{\sin t}{t}e^{-st}\,dt\tag{1}$$ fulfills (by the dominated convergence theorem) $$ g'(s) = -\int_{0}^{+\infty}\sin(t)e^{-st}\,dt = -\frac{1}{1+s^2}\tag{2} $$ hence $$ g(s) = C-\arctan(s) \tag{3} $$ for some constant $C$. Since, by (1) and the fact that $\frac{\sin t}{t}$ is a bounded smooth function, it is obvious that $\lim_{s\to +\infty}g(s)=0$, such a constant is $C=\frac{\pi}{2}$ and $$ \mathcal{L}\left(\frac{\sin t}{t}\right)(s) = \color{red}{\arctan\left(\frac{1}{s}\right)}.\tag{4} $$

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  • $\begingroup$ @KohJingYu: Your proposed edit would have been better handled as a Comment to the author of this Answer. Keep participating in Math.SE and you will soon earn the reputation needed for the privilege of commenting on the posts of others. $\endgroup$
    – hardmath
    Apr 7 '17 at 14:23

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