1

Here, I saw the following formula:

$$\int_{0}^{\infty }\frac{f(t)}{t}dt=\int_{0}^{\infty }\mathcal{L}\left \{ f(t) \right \}ds$$

Say we have the integrand only $f(t)$, not $\frac{f(t)}{t}$, then I think we can multiply and divide this integrand by $t$, then we apply the formula. For instance, if the integrand is $te^{-t}$, then we should rewrite it as

$$\frac{t^2e^{-t}}{t}$$

So $f(t)$ is actually $t^2e^{-t}$, not $te^{-t}$.

I know that if the integrand is $te^{-t}$, then it can be done by integrating by part. However, I think it is a good looking function to illustrate my question.

Am I right about multiplying and dividing by $t$ just to make a use of the formula?

In other words, is the following true?

$$\int_{0}^{\infty }f(t)dt=\int_{0}^{\infty }\mathcal{L}\left \{ tf(t) \right \}ds$$

Second Question:

Can we, somehow, generalize the formula to any interval $(a,b)$ instead of $(0,\infty)$?

I know that the Laplace will have the same limits of integration, $(0,\infty)$ but I am asking about the original integral, not the integral from the $\mathcal{L}$.

Sorry for my bad English, hope my questions are clear. Your help would be appreciated. THANKS!

Edit:

For the second question, what I mean is

Given $a,b,$ and $h(x)$, and that $\int_{a}^{b}h(x)dx = \int_{0}^{\infty}r(t)dt$. How to find $r(t)$?

This will make use of the first formula in this post.

For instance, $\int_{1}^{2}\sqrt{4x(2-x)}dx$ can be replaced by $\int_{0}^{\infty} \frac{dt}{1+t^2}$. And now we can use the above formula.

Hussain-Alqatari
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    In general, $$\int_{0}^{\infty} f(x) \cdot g(x) , dx = \int_{0}^{\infty} {\mathcal{L} f}(s)\cdot{\mathcal{L}^{-1} g}(s) , ds$$ Set $g(x) = \frac{1}{x}$ and this yields your first formula. There are no such symmetries for an arbitrary interval $(a,b)$. – KStarGamer Jun 29 '22 at 12:43

2 Answers2

1

For your new question , use the transformation $y=\frac{x-a}{b-x}$ and then $x=\frac{yb+a}{1+y}$ and $dx=\frac{(b-a)dy}{(1+y)^{2}}$. Thus the l.h.s integral can be written as $\int_{0}^{\infty }h(\frac{yb+a}{1+y})(1+y)^{-2}(b-a)dy$. therefore you have the $r(y)$ you wanted!!

0

Write the r.h.s integral as $\int_{0}^{\infty }\int_{0}^{\infty }tf(t)e^{-st}dtds$ and change the order of integration. Then

$\int_{0}^{\infty }$$tf(t)e^{-st}ds=tf(t)(1/t)=f(t)$ and hence your claim is true!