Does this integral converge or diverge:
ʃ(8lnx)/(x^8)
where a = 1 and b = infinity
(hint: integrate by parts).
ʃ(8lnx)/(x^8)
where a = 1 and b = infinity
(hint: integrate by parts).
-
ʃ(8lnx)/(x^8)
Let t = ln x
dt = 1/x dx
ʃ(8lnx)/(x^8) dx = 8 ʃ t /x^7 dt =
ln x = t
x = e^t
x^7 = e^7t
8 ʃ t /x^7 dt = 8 ʃ t e^(-7t) dt
When x=1 t=ln 1 = 0
when x approaches infinity t = ln x approaches infinity as well
we need to evaluate 8 ʃ t e^(-7t) dt from 0 to infinity
Integrate by parts
dv=e^-7t
v=-e^-7t / 7
u=t
du = dt
ʃ u dv = u v - ʃ v du
8 ʃ t e^(-7t) dt = -t e^-(7t) /7 + (1/7) ʃ e^(-7t) dt
= (-8/49) e^(-7t)
Let F(t) = (-8/49) e^(-7t)
F(infinity) = 0
F(0) = -8/49
0-(-8/49) = 8/49
The integral is 'proper' and converges.
Let t = ln x
dt = 1/x dx
ʃ(8lnx)/(x^8) dx = 8 ʃ t /x^7 dt =
ln x = t
x = e^t
x^7 = e^7t
8 ʃ t /x^7 dt = 8 ʃ t e^(-7t) dt
When x=1 t=ln 1 = 0
when x approaches infinity t = ln x approaches infinity as well
we need to evaluate 8 ʃ t e^(-7t) dt from 0 to infinity
Integrate by parts
dv=e^-7t
v=-e^-7t / 7
u=t
du = dt
ʃ u dv = u v - ʃ v du
8 ʃ t e^(-7t) dt = -t e^-(7t) /7 + (1/7) ʃ e^(-7t) dt
= (-8/49) e^(-7t)
Let F(t) = (-8/49) e^(-7t)
F(infinity) = 0
F(0) = -8/49
0-(-8/49) = 8/49
The integral is 'proper' and converges.