[Discussion] Solving the Limit Problem

[yangyongkang]

Nonsense, paste code

Code:

limit((∫(∫(sin(t)*atan(1+t),t,0,u^2),u,0,x))/(x^3*((x+1)^(1/3)-1)^2),x,0)

xcas get

Code:

"Limit: Max order reached or unable to make series expansion Error: Bad Argument Value"

mathematica also calculated for a long time

Code:

Limit[Integrate[Sin[t]*ArcTan[1 + t], {u, 0, x}, {t, 0, u^2}]/(
 x^3*((x + 1)^(1/3) - 1)^2), x -> 0]

Surprisingly, the Ti Nspire CX CII CAS is calculated

In fact, the hp prime can be calculated, and it needs to be replaced by another method.

let f(t)=sin(t)*atan(1+t)

Code:

series((∫(∫(f(t),t,0,u^2),u,0,x)/(x^3*((x+1)^(1/3)-1)^2)),equal(x,0),1)

hp prime get

Code:

(3*f(0)/x^2)+(2*f(0)/x)+(9/10)*(function_diff(f))(0)-(1/9)*f(0)+((3/5)*(function_diff(f))(0)+(2/27)*f(0))*x+x^2*order_size(x)

We find f (0) = 0 and substitute it into the result,Get the limit

Code:

((9/10)*∂(sin(t)*atan(1+t),t)|(equal(t,0)))

hp prime get

Code:

9*π/40

This is the correct answer

[yangyongkang]

Several perverted integrals
(i)

Code:

evalf(int(1/(1+e^(1/x)),x,-1,1))

The answer is 1
xcas get

Code:

"Unidirectional limits are distincts 1,0 Error: Bad Argument Valu"

(ii)

Code:

int(x^2*tan(x)^2+x^2+2*x*tan(x)+1,x)

The answer is

Code:

x+x*tan(x)

(iii)

Code:

simplify(int((x*tan(x)+ln(x*cos(x))-1)/ln(x*cos(x))^2,x))

The answer is

Code:

x/ln(x*ln(x))