Laplace Transformations of $\frac{1}{t}$

Question

what is the laplace transform of $\frac{1}{t}$? I tried different ways like integrating by parts from the general form of laplace but it's getting more complex as my solution goes by.

Answer 1

I may help you with that:

I was working with those kind of function today (1/X) and (1/X). And I must say that the we must remember one of the first transformation we did.

Remember that $L\left\{ H(x) \right\} =\frac { 1 }{ s }$ where $H(x)$ is the heavisde function, then If that is truth, the next thing to do is to write the anti-laplace transform definition: $$H(x)=\int _{ 0 }^{ \infty }{ \frac { 1 }{ s } } { e }^{ sx }ds$$ Well, this expression is "very" similar to the one that we are trying to figure out, and it gets better, we only have to make two more steps to get the expression of the laplace transform of (1/X): First: lest make a doble change of vars: where there is a X we will put a S, and where there is a S we will put a X.

The resultant expression is: $$H(s)=\int _{ 0 }^{ \infty }{ \frac { 1 }{ x } } { e }^{ sx }dx$$ Second, where there is a S we will put -S, the resultan expression is: $$H(-s)=\int _{ 0 }^{ \infty }{ \frac { 1 }{ x } } { e }^{ -sx }dx$$ This means that the $$H(-s)=L\left(\frac { 1 }{ x } \right)$$

$H(-s)$ is the function that is equally to $1$ from $-\infty$ to $s=0$, from $s=0$ to $\infty$ the function is NULL.