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Question: why putting assumptions inside simplify gives an error but outside it works?

Question:why putting assumptions inside simplify gives an error but outside it works?

nm 10229 Maple
simplify assume
December 27 2021
1

I do not understand why  simplify(eq,size,assume =t::real); gives an error but simplify(eq,size) assuming t::real; does not.

Which is the correct way to use assumptions with simplify? Inside or outside? And why would it make a difference?

Maple 2020.2 on windows 10.

> 

interface(version);

`Standard Worksheet Interface, Maple 2021.2, Windows 10, November 23 2021 Build ID 1576349`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1122 and is the same as the version installed in this computer, created 2021, December 22, 16:3 hours Pacific Time.`

> 

restart
> 

eq:=t = 1/2*Int(1/ln(exp(t^2-2*t*y+y^2))^(1/2)/exp(t^2-2*t*y+y^2)/(exp(t^2-2*t*y+y^2)-1),exp(t^2-2*t*y+y^2))+c[1];
simplify(eq,size,assume =t::real);
 

t = (1/2)*(Int(1/(ln(exp(t^2-2*t*y+y^2))^(1/2)*exp(t^2-2*t*y+y^2)*(exp(t^2-2*t*y+y^2)-1)), exp(t^2-2*t*y+y^2)))+c[1]

Error, (in assuming) when calling '`anonymous procedure called from tools/recurse/indets`'. Received: 'invalid input: `simplify/int/simplify` expects its 3rd argument, r, to be of type {name, list({range, name = range}), name = anything}, but received exp((t-y)^2)'

> 

simplify(eq,size) assuming t::real;

t = (1/2)*(Int(1/(ln(exp((y-t)^2))^(1/2)*exp((y-t)^2)*(exp((y-t)^2)-1)), exp((y-t)^2)))+c[1]

> 

 

I know the above inert int looks strange, but this is why it is inert. Later on there will be change of variable to make the integration variable a single symbol again as in normal integration usage.

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