Given that the function f (x) is continuous on (- ∞, + ∞) and satisfies ∫ (0, x) f (x-u) e ^ UDU = SiNx, X ∈ (- ∞, + ∞), find f (x)

Given that the function f (x) is continuous on (- ∞, + ∞) and satisfies ∫ (0, x) f (x-u) e ^ UDU = SiNx, X ∈ (- ∞, + ∞), find f (x)

Let x-u = t, Du = - DT
u=0,t=x
u=x,t=0
∫[0,x] f(x-u)e^udu
=∫[x,0] f(t)e^(x-t)*(-dt)
=∫[0,x] f(t)e^(x-t)dt
=e^x∫[0,x] f(t)e^(-t)dt
=sinx
∫[0,x] f(t)e^(-t)dt=sinx/e^x
The derivation on both sides is very good
f(x)e^(-x)=(sinx/e^x)'
=(cosxe^x-sinxe^x)/(e^x)^2
=(cosx-sinx)/e^x
f(x)=cosx-sinx

Is the function e ^ (- x ^ 2) * SiNx infinitesimal when x tends to infinity?

A:
-1

When x → 0, the infinitesimal order of e ^ (x ^ 4-x ^ 3) - 1 (process required)

When x → 0, because e ^ X-1 ~ x, so
e^(x^4-x^3)-1 (x^4-x^3)
The order of ^ x-3 can be obtained
The order is 3