Laplace transforms for which value of s?

Haku
Messages
30
Reaction score
1
Homework Statement
In each case, state the values of s for which the
Laplace transform exists.
Relevant Equations
Laplace transform
I was wondering how you work out what values of s a Laplace transform exists? And what it actually means? The example given in class is an easy one and asks to calculate the Laplace transform of 3, = 3 * Laplace transform of 1 = 3 * 1/s. Showing this via the definition, where does the range of s come out of? I.e. how can I define when the Laplace transform of 3 = 3/s?
And what happens outside that range?
Thanks.
 
Physics news on Phys.org
I do not read the original question so as a general observation, s>0 because factor ##e^{-st}## in Laplace integral should diminish for large t>0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top