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

Alec_

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Hello,

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

[math]\int\dfrac{1}{s(t)} ds[/math]
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.
 
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 [imath]s(t)[/imath] one cannot compute this integral. It's even worse: for some known [imath]s(t)[/imath] the integral might not have a "closed form" expression.
 
Hello,

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

[math]\int\dfrac{1}{s(t)} ds[/math]
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:

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

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

[math]\int\dfrac{1}{s(t)} ds[/math]
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.
 
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
[imath]\displaystyle \int \dfrac{ds}{s(t)} = ln|s(t)| + f(t)[/imath]

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

-Dan
 
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