Laplace transform for 1/ x?

By Andrew Hodges on Saturday, September 08, 2001 - 10:15 pm:

I don't think there is a Laplace transformation for either the function 1/x or ln x and I think I have a proof of this. Does this check out?

Let 1/x = 1/(1 - y) These functions are in principle the same, i.e. f(x) = f(1 - y)

Also, 1/(1 - y) = 1 + y + y² + y³ + y⁴ .....

(for |y| < 1)

Hence L(1/(1 - y) ) = L(1 + y + y² .......)

Letting L(y) = F(s) we get
L(1 + y + y² +.....) = 0!/s + 1!/s² + 2!/s³ + 3!/s⁴ .............

This is the sum from 1 to infinity of (n-1)!/sⁿ

However, if we apply the ratio test, we obtain the result that this series has a radius of convergence of infinity, and so it cannot converge, and so therefore there cannot be a Laplace transformation for 1/x.

The argument for ln x is similar; we let x = 1-y and take the power series for ln(1-y)

= 1 + y + y² /2 + y³ /3 + y⁴ /4 +......

We then carry out the Laplace transform on that series again, and find that it also has a radius of convergence of infinity, and so cannot have a Laplace transform. I am worried about a couple of things about this proof though, for instance it only applies when mod y< 1; could there be some range of values for which there is a Laplace transform for 1/x or ln x?

By David Loeffler on Monday, September 10, 2001 - 10:37 am:

I'm afraid that when you say that the functions are ''in principle the same'', this is not quite true. In fact the definition of the Laplace transform is

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

, so the range of integration is important. Readjusting the variables as you have messed up the limits; you were, in effect, integrating from

- \infty

to 1.

Also, isn't this a rather roundabout method? All that is needed is to show that

$\int_{0}^{\infty} e^{- s t} / t dt$

doesn't exist; and this is not difficult if you consider its series expansion at the origin. At the origin it is approximately $1 / t - s + 1 / 2 s^{2} t$ and so on. The remaining terms are all integrable on [0,1] and form a convergent series (prove this if you want). The first term's integral on [0,1] diverges. Hence $\int_{0}^{1} e^{- s t} / t dt$ doesn't exist, so the infinite integral doesn't either.

David