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How can I solve this integral equation using Laplace transform?

$${\int\limits_0^{\infty}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname d\!t$$

Knowing that $$ \mathcal{L}\{\cos t\} = \frac{s}{s^2+1} $$

I think I can start by taking limits:

$$\lim_{b \rightarrow \infty} {\int\limits_0^{b}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname d\!t$$

ant then apply the shortcut of $$\mathcal{L}\{\cos t\}$$

but I don't know how to continue. Any help will be appreciated.

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  • $\begingroup$ Where is the equation? I cannot see any equal sign in your post. $\endgroup$ – enzotib May 10 '14 at 10:11
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We can use the property $$ \mathcal{L}\left[\frac{f(t)}{t}\right](s)=\int_s^\infty \mathcal{L}[f(t)](s')ds' $$ so we have $$ \int_0^\infty\frac{1-\cos t}{t}e^{-t}dt =\mathcal{L}\left[\frac{1-\cos t}{t}\right](1)=\\ \int_1^\infty\mathcal{L}[1-\cos t](s)ds =\int_1^\infty\left(\frac{1}{s}-\frac{s}{1+s^2}\right)ds\\ \left.\left(\log s-\frac{1}{2} \log(1+s^2)\right)\right|_{1}^\infty =\left.\log\frac{s}{\sqrt{(1+s^2)}}\right|_{1}^\infty=-\log\left(\frac{1}{\sqrt{2}}\right) $$

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To be honest I'm not sure what you meant by 'apply the shortcut of ...', but one way to do it is by writing $$1/t = \int_0^{\infty}e^{-tx}\, dx $$ so that $$\int_0^{\infty} \frac{e^{-t}(1-\cos t)}{t} dt = \int_0^{\infty} e^{-t}(1-\cos t) \int_0^{\infty} e^{-tx} \, dx \, dt \\ = \int_0^{\infty} \int_0^{\infty} e^{-(x+1)t}(1-\cos t) \, dt \, dx$$ Then apply your knowledge of the Laplace transform of $\cos$ to calculate the inner integral. The outer integral can then be evaluated by elementary methods.

In somewhat more detail, you obtain $$ \int_0^{\infty}\frac{1}{x+1}-\frac{x+1}{1+(x+1)^2} \, dx $$ which you can integrate using the primitive $$\ln \left[ \frac{x+1}{\sqrt{1+(x+1)^2}} \right]$$

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  • $\begingroup$ I don't know either you realized it or not but you actually use a similar approach as enzotib's answer. I give +1 for both of you. Very nice answers! :) Anyway, I have another way to solve this integral but I'm too lazy to write the answer since this question get less viewers. :D $\endgroup$ – Tunk-Fey May 10 '14 at 14:08

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