How to understand the Laplace transform of $f(t)/t^2$

Question

I am trying to solve certain type of differential equations. For this I was trying to obtain $\mathcal{L}\{f(t)/t^2\} $. There is a question related to this for another similar case when f(t)/t, at here Laplace transform of $f(t)/t$

According to the book Tables of Laplace transforms. Springer Science & Business Media. Oberhettinger (2012), page 4. Such Laplace transform exist and is represented in the general form.

$\mathcal{L}\{t^{-n}f(t)\} = \int_p^\infty ... \int_p^\infty g(p) (dp)^{n}$

for my particular case:

$\mathcal{L}\{f(t)/t^2\} = \int_p^\infty \int_p^\infty g(p) (dp)^{2}$

It is still unclear to me, what is the purpose of having this kind of "dummy" multidimensional integral. And why it would help in the process to make things "better" so I can obtain the inverse Laplace transform of my nonlinear ode.

Thanks for your comments!

Ps. The nonlinear ODE is the following:

$\left(\frac{A}{r^2}+B r^2+ C \right) u(r)-u''(r)=0$

Answer 1

If $f$ is piecewise continuous on $[0,+\infty[$ and of exponential order $\alpha$ and assume $\lim_{t\to 0^{+}}(f(t)/t)$ exists, so $$L\left\{ \frac{f(t)}{t}\right\}(s)=\int_{s}^{+\infty}F(u)\, {\rm d}u,$$ where $F(s):=L\{f\}(s)$ it is follows since:

• $\frac{d}{ds}L\left\{\frac{f(t)}{t}\right\}(s)=-F(s)$,
• $|f(t)|\leqslant Me^{\alpha t}$ for all $t\geqslant 0$,
• $\lim_{s\to +\infty}L\{f\}(s)=0$.

Then, substitution $f(t)$ by a new function $f(t)/t$ and then you have $$L\left\{\frac{f(t)/t}{t}\right\}(s)=L\left\{\frac{f(t)}{t^{2}}\right\}(s)=\int_{s}^{+\infty}L\left\{\frac{f(t)}{t} \right\}(u)\, {\rm d}u=\int_{s}^{+\infty}\int_{r}^{+\infty}F(v) \, {\rm d}v{\rm d}u.$$ Then, by mathematical induction over the $n$ natural you should arrive to $L\{f(t)/t^{n}\}$ as in your textbook.

N.B: I don't know anything about your non-linear ODE, so I don't know how the identity shown can be useful, so I can't answer that part.