Indefinite integral: integral [ 1/s(t) ] ds

[Alec_]

Hello,

I am currently working on a university coursework but I am struggling on one step:
How do I integrate:

$$\int\dfrac{1}{s(t)} ds$$
We are not given an expression of s(t).
For context, this is s(t) from the SIR disease propagation model.

Thank you for your help.

 

[blamocur]

For context, this is s(t) from the SIR disease propagation model.

Most of us have no clue what "SIR disease propagation model." is. Without knowing $s(t)$ one cannot compute this integral. It's even worse: for some known $s(t)$ the integral might not have a "closed form" expression.

 

[khansaheb]

Hello,

I am currently working on a university coursework but I am struggling on one step:
How do I integrate:

$$\int\dfrac{1}{s(t)} ds$$
We are not given an expression of s(t).
For context, this is s(t) from the SIR disease propagation model.

Thank you for your help.

As I see it:

$\displaystyle \int{\frac{ds}{s}} = ln(s) + c$

 

[BobP]

Hello,

I am currently working on a university coursework but I am struggling on one step:
How do I integrate:

$$\int\dfrac{1}{s(t)} ds$$
We are not given an expression of s(t).
For context, this is s(t) from the SIR disease propagation model.

Thank you for your help.

The notation s(t) says that s is a function of t. To carry out the integration you need to know what the relationship is between s and t.

 

[topsquark]

The notation s(t) says that s is a function of t. To carry out the integration you need to know what the relationship is between s and t.

Not quite. We are integrating over s, not t. So
$\displaystyle \int \dfrac{ds}{s(t)} = ln|s(t)| + f(t)$

where f(t) is an arbitrary function of t. But, yes, to get anywhere further, we need to know more.

-Dan