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I am taking a differential equation class and for Laplace transformations and I have to find $$\displaystyle \int_0^\infty \dfrac{\sin t}{t}dt.$$

How can I do that?

Amzoti
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madU
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1 Answers1

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We know that $$\int_0^\infty \frac{f(x)}{x} \, dx=\int_0^\infty F(s) \, ds$$ (memorize it). Now setting $f(x)=\sin x$ and $\mathcal{L}\{\sin x\}=\dfrac{1}{s^2+1}$, we have $$\int_0^\infty \frac{\sin x}{x} \, dx=\int_0^\infty \frac{1}{s^2+1} \, ds=\arctan s\Big|_0^\infty =\pi/2$$

Michael Hardy
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Mikasa
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  • tnx, but i didn't know the formula – madU Aug 24 '13 at 17:43
  • we were only given L{f(t)/t} = integral F(s) ds from 0 to infinity – madU Aug 24 '13 at 17:44
  • could you please mention from where you took ∫∞0f(x)/xdx=∫∞0F(s)ds – madU Aug 24 '13 at 17:53
  • @madU: Theorem: If $\mathcal{L}{f(x)}=F(s)$ and if the L.T of $g(x)=\frac{f(x)}{x}$ exists then we have the above formula. – Mikasa Aug 24 '13 at 17:59
  • ok, i still didn't get it completely, but tnx a lot i'll try to go through it and understand the theorem you mentioned – madU Aug 24 '13 at 18:01
  • @madU: Follow this link – Mikasa Aug 24 '13 at 18:14
  • $+1$ for my dear friend! – amWhy Aug 25 '13 at 00:33
  • The integral $\int_0^\infty \dfrac{\sin x}{x},dx$ converges conditionally, so the usual Fubini's theorem doesn't justify the change in order of integration that gives you the identity that this answer tells the reader to memorize. Instead, one has to write $\lim_{a\to\infty}\int_0^a \dfrac{\sin x}{x},dx$, then apply Fubini's theorem, and then take the limit as $a\to\infty$ only after that. So there are some complications. – Michael Hardy Aug 25 '13 at 02:50
  • tnx for all, finally i got it – madU Aug 25 '13 at 02:52