# Laplace transform

• Dec 26th 2010, 06:20 AM
vishnuk5320
Laplace transform
Why does laplace trancsform of trigonometric functions like tan,sec , cosec,cot,..and logarithmic functions doesnot exist? what is the reason behind that?
pls help friends...
• Dec 26th 2010, 07:16 AM
chisigma
The question has to be proposed in a more precise form!... for example the 'logarithmic function' $\ln t$ is perfectly L-transformable...

Kind regards

$\chi$ $\sigma$
• Dec 26th 2010, 07:41 AM
chisigma
... in general necerssary condition for the existence of the integral...

$\displaystyle \int_{0}^{\infty} f(t)\ e^{-s t}\ dt$ (1)

... is that $f(*)$ is locally integrable on $[0,\infty)$. A function is said to be 'locally integrable' if it is integrable on any compact set of its domain of definition. Functions like $\displaystyle \tan t$, $\displaystyle \frac{1}{\sin t}$ , $\displaystyle \frac{1}{t}$, $\displaystyle \frac{1}{t^{2}}$ etc... are not locally integrable on $[0,\infty)$ ...

Kind regards

$\chi$ $\sigma$
• Dec 27th 2010, 12:39 AM
vishnuk5320
i mean L(log t).....
• Dec 27th 2010, 02:10 AM
FernandoRevilla
Quote:

Originally Posted by vishnuk5320
i mean L(log t).....

Use the integral expression of the Euler Mascheroni constant:

$\gamma=-\displaystyle\int_0^{+\infty}e^{-x}\log x\;dx$

you'll obtain:

$\mathcal{L}\{\log t\}=-\dfrac{\log s +\gamma}{s}$

Fernando Revilla
• Dec 27th 2010, 10:55 AM
FernandoRevilla
Maybe I missed your initial question:

Quote:

Originally Posted by vishnuk5320
Why does laplace trancsform of trigonometric functions like tan,sec , cosec,cot,..and logarithmic functions doesnot exist? what is the reason behind that?

Perhaps you mean that you don't see some Laplace transformed in the standard tables. The reason is that some of them, don't exist (it has been already commented by chisigma). In other cases, (for example, for the logarithmic funcion) we need some previous results.

Fernando Revilla