this seems like a homework problem. yes! To some extent. But really I was not getting it.

I was not able to get the Laplace transform of $\cos(t)/t$. using the property of Integration in Laplace Transform,

 Integration in Laplace Transform

but since the integral is not convergent, I am not able to get the laplace transform.


but I know it exist!

then how to get the Laplace Transform?

how can It be possible that Using one Method you are getting and other (Integration method) you are not getting the answer ?

  • 3
    $\begingroup$ Part of it has to do with the fact that $\cos(0)\neq 0$ so the Laplace transform existing in the first place from definition doesn't seem likely since that integral diverges too $\endgroup$ – Ninad Munshi Aug 17 '20 at 5:10
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    $\begingroup$ The Laplace transform for your function is $$\mathcal{L}\{f\} = \int_0^\infty \frac{\cos(t)e^{-st}}{t} \, dt,$$ near $t=0$ this integral doesn't converge since $\big($using $\cos(x)\approx 1 + O(x^2)$ by the Maclaurin expansion$\big)$ $$\lim_{\varepsilon \to 0^+}\int_0^\varepsilon \frac{\cos(t)e^{-st}}{t} \, dt\approx \lim_{\varepsilon \to 0^+}\int_0^\varepsilon\frac{1}{t}\,dt$$ is divergent. Even though the integral diverges, the Laplace transform still exists in the sense of distributions. However, I am guessing that your homework was to observe that the integral diverges? $\endgroup$ – Axion004 Aug 17 '20 at 17:07

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