已知∫(x,0) f(t-n)e^n dt=sinx,求f(x)

已知∫(x,0) f(t-n)e^n dt=sinx,求f(x)
已知∫(x,0) f(t-n)e^n dt=sinx,求f(x)

已知∫(x,0) f(t-n)e^n dt=sinx,求f(x)
∫(0→x) f(t - n)e^n dt = sinx
f(x - n)e^n = cosx
f(x - n) = (cosx)/e^n
f[(x + n) - n] = cos(x + n)/e^n
f(x) = e^(- n)cos(x + n)

设t-n=y，
则左边=∫(0→x)f(y)e^ndy=e^n∫(0→x)f(y)dy=sinx
所以f(x)=-cosx/e^n