# Does the Laplace Transform of e^(at)/t Exist?

• patric44
In summary, the conversation discusses the Laplace transform of a function and the use of a theorem to calculate it. However, due to certain requirements of the theorem, the Laplace transform in this case does not exist. There is a discrepancy between the calculated transform and the expected result, leading to a discussion of the validity of the theorem.
patric44
Homework Statement
find Laplace transform of e^(at)/t
Relevant Equations
L(e^at/t)
hi guys
i am facing a little problem calculating this Laplace transform ## \mathscr{L}(\frac{e^{\alpha t}}{t})## , when calculate it using the method of the inverse Laplace transform its equal to
$$ln{\frac{1}{s-\alpha}}$$
but then when i try to use the theorem
$$\mathscr{L}(\frac{f(t)}{t})=\int_{s}^{\infty}F(s)ds=\int_{s}^{\infty}\mathscr{L}(f(t))ds = \int_{s}^{\infty}\frac{1}{s-\alpha}ds$$
$$=lim_{s→∞}(s-\alpha)-ln(|s-\alpha|)$$
it seems that there is a term that will blow up to infinity!
what i am missing here?!

Last edited by a moderator:
I think you should start by writing down the integral that is the definition of the laplace transform, and ask yourself if it exists.

Office_Shredder said:
I think you should start by writing down the integral that is the definition of the laplace transform, and ask yourself if it exists.
i know that it will lead me to the exponential integral function Ei(x) , but its easy to show that the ##L^{-1}(ln\frac{1}{s-\alpha}) = e^{\alpha t}/t##

patric44 said:
i know that it will lead me to the exponential integral function Ei(x) , but its easy to show that the ##L^{-1}(ln\frac{1}{s-\alpha}) = e^{\alpha t}/t##
The assumption in the Laplace transform is that ##f## in the integral ##\int_0^{\infty} f(t)e^{-st}dt## must be one that is well behaved, in the sense that it is at least conditionally convergent at infinity.

I think the laplace transform of this function simply does not exist because the integral $$\int_0^{+\infty}\frac{e^{-(s-a)t}}{t}dt$$ does not exist, because for example the integral $$\int_0^1\frac{e^{-(s-a)t}}{t}dt$$ does not exist and the "rest" integral $$\int_1^{+\infty}\frac{e^{-(s-a)t}}{t}dt$$ is simply positive.

That theorem doesn't hold because the theorem has some requirements (at least according to my book which is written in Greek by Greek authors) and one of the requirements is the limit $$\lim_{t\to 0}\frac{f(t)}{t}$$ to be a real number, which in this case doesn't hold.

jbunniii

## 1. What is the Laplace transform of exp(at)/t?

The Laplace transform of exp(at)/t is given by the function F(s) = 1/(s-a) where s is the complex variable and a is a constant.

## 2. What is the significance of the Laplace transform of exp(at)/t?

The Laplace transform of exp(at)/t is used to solve differential equations in engineering and physics. It is also a useful tool in signal processing and control systems.

## 3. How is the Laplace transform of exp(at)/t calculated?

The Laplace transform of exp(at)/t is calculated using the formula F(s) = ∫[0,∞] exp(-st) * exp(at)/t dt, where ∫[0,∞] represents the integral from 0 to infinity.

## 4. What is the inverse Laplace transform of exp(at)/t?

The inverse Laplace transform of exp(at)/t is given by the function f(t) = exp(at). This means that the original function exp(at)/t can be obtained by taking the Laplace transform of exp(at).

## 5. What are the applications of the Laplace transform of exp(at)/t?

The Laplace transform of exp(at)/t has various applications in engineering, physics, and mathematics. It is used to solve differential equations, analyze control systems, and study signals in communication systems. It is also used in probability and statistics to calculate moments of random variables.

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