Convergence/divergence of an improper integral?

Question

How to determine whether the following are convergent or divergent:

Integral (upper:2, lower:1) dx/ x ln x

Integral (upper 3, lower: 0) 1/(sqrt(x)) e^(-sqrt(x)) dx

If convergent, how to find the value?

Answer 1

Integral (1 to 2, (1/(xln(x)) dx )

Since the function 1/(xln(x)) will become undefined when x = 1, we must convert this limit accordingly by taking the limit as t approaches from the right of 1.

lim Integral (t to 2, 1/(xln(x)) dx )
t -> 1+

lim Integral (t to 2, 1/(ln(x)) (1/x) dx )
t -> 1+

Use substitution.
Let u = ln(x). [when x = t, u = ln(t). When x = 2, u = ln(2).]
du = (1/x) dx.

lim Integral (ln(t) to ln(2), 1/u du)
t -> 1+

lim [ln|u|] {evaluated from ln(t) to ln(2)}
t -> 1+

lim ln(ln(t)) - ln(ln(2))
t -> 1+

As t approaches 1+ from the right, ln(t) approaches negative infinity, which means ln(ln(t)) approaches negative infinity.
ln(ln(2)) is a finite number, so we are adding negative infinity to a finite number.

Therefore, this diverges.

Answer 2

Convergence is normally checked at infinity. Since both limits are known we can actually find the value of the integral
In first case, assuming lnx in the denominator we get 'lnlnx 'as the answer which gives us lnln2 - lnln1. In this case lnln1= ln0 = -infinity.
Actually we r calculating the limit of lnlnx at x=1.
So the integral is lnln2+infinity=infinity. Hence it diverges.
As for the second one;
Assuming the exponent in the numerator (next time use brackets). The integral is '-2e^(-sqrt(x)) '. So within the limits it will be -2(e^(-sqrt3) - e^(-sqrt0))=2(1-e^(-sqrt3)) which is a finite quantity and converges at this value

Answer 3

i) essential (0 to infinity) tanh(ln x) dx = essential (0 to infinity) [e^(ln x) - e^(-ln x)]dx/ [e^(ln x) + e^(-ln x)] = essential (0 to infinity) [x - x^(-a million)]dx/ [x + x^(-a million)] = essential (0 to infinity) (x^2 - a million) dx/ (x^2 + a million) = essential (0 to infinity) [a million - 2/ (x^2 + a million)] dx = x - 2arctan(x) from 0 to infinity = (infinity - pi) - (0) Diverges to infinity. ii) enable I = essential (0 to pi/2) cos x /sqrt(sin x) dx considering that y > sqrt(y) for y in (0,a million), a million/sin y < a million/sqrt(sin y). as a outcome, I >= essential (0 to pi/2) cos x /sin x dx = ln|sin x| from 0 to pi/2 = 0 - (-infinity) = infinity. So, by ability of the evaluation attempt, the unique unsuitable essential I diverges.