Question

\[\mathcal{L}\Big\{{1-cos(ax) \over x }\Big\}\]
use
\[\mathcal{L}\Big\{{f(x) \over x}\Big\}=\int_s^∞\mathcal{L}\Big\{ f(x)\Big\} ds \]

Answer 1

Twice I started to reply, thinking I could solve it. Twice I was wrong.

XP

Answer 2

I've never seen that identity before, and it makes me a little uncomfortable. The laplace transform of a function should be a function of the complex variable s, but the expression on the right is just a number...

Answer 3

I know this isn't particularly helpful, but I want to see a reference on my screen, so just posting it.

Answer 4

Oh, man. The integration bounds are so small, I misread. Also as a general rule, when you write integrals, you definitely don't want the integration variable anywhere near the bounds. It would be better to say
\[ L\{ \frac{f(x)}{x} \} = \int_s^\infty F(\sigma) d\sigma \]

Answer 5

The Laplace transform of the numerator is
\[\frac{1}{s} - \frac{s}{s^2 + a^2} \]
So we can integrate to find:
\[\int_s^b \frac{1}{\sigma} + \frac{\sigma}{\sigma^2 + a^2} d\sigma \]
\[ = \log(\sigma) - \frac{1}{2}\log(\sigma^2 + a^2) |^b_s \]
\[ = \log\left( \frac{\sigma }{\sqrt{\sigma^2 + a^2}} \right)|^b_s \]
\[ = \log \left( \frac{b}{\sqrt{b^2+a^2}} \right) - \log \left(\frac{s}{\sqrt{s^2 + a^2}}\right)\]
taking the limit as b - > infinity, the first term becomes log(1) = 0, leaving us with our answer,
\[ -\log\left( \frac{ s}{\sqrt{s^2 + a^2} } \right) = \log\left( \frac{\sqrt{s^2+a^2}}{s}\right) \]
Or hopefully something of that sort.... haha

Answer 6

The answer in the back of my book is
\[ln{\sqrt{1+{a^2 \over s^2}}}\]

Answer 7

Awesome! I was right then haha

Answer 8

Because of course,
\[ \log\left(\frac{\sqrt{s^2 + a^2}}{s} \right) = \log \sqrt{1 + \frac{a^2}{s^2}} \]

Answer 9

Excuse the dumb question, why did we take the limit as b went to infinity? In your solution?

Answer 10

hey good stuff , i havn't had time to go over your working yet , and my laptop is about to run outta battery
but thank youse so much

Answer 11

@badreferences
b?

Answer 12

Oh, nevermind, lol, I see. Forgot the original question.

Answer 13

oh there is a b i couldn't see it right away,
i havent studied the solution yet but i can see it is correct

Answer 14

I just used the definition of an improper integral:
\[ \int_0^\infty f(x) dx = \lim_{b \rightarrow \infty} \int_0^b f(x) dx \]

Answer 15

Yeah, I saw. I tend to lose sight of the original problem when I'm doing solutions.

Answer 16

\[\mathcal{L}\Big\{{1-cos(ax) \over x }\Big\}(s)\]\[=\int_s^∞\mathcal{L}\Big\{1-cos(ax)\Big\}(s)ds\]\[=\int_s^∞({1 \over \sigma}-{{\sigma \over \sigma^2+a^2}})ds\]\[=ln(\sigma)-{1 \over 2}ln(\sigma^2+a^2)\Big|_s^∞\]\[=ln({\sigma \over \sigma^2+a^2})\Big|_s^∞\]\[=ln(1)-ln({s \over \sqrt{s^2+a^2}})\]\[=ln(\sqrt{s^2+a^2 \over s})\]\[=ln(\sqrt{1+{a^2 \over s^2}})\]

Answer 17

i was trying to use one of these integrals from a table

\[\int{d\sigma \over \sigma^2+a^2}={1 \over a}tan^{-1}({\sigma \over a})+c\]
\[\int{d\sigma \over \sigma^2-a^2}={1 \over 2a} ln|{\sigma+a \over \sigma-a}|\]


Totally does not work