Is the improper integral convergent or divergent?: A.) (ln(x+1))/x^2 dx (with lower bound, 0 and upper bound, 1) B.) (ln(x+1))/x^2 dx (with lower bound, 1 and upper bound, infinity)

Answer 1

Jacques gives a great answer.

Another way of looking at this is to do comparison tests.

ln(x+1) -> x as x goes to 0.

Thus, the function ln(1+x)/x^2 goes to 1/x as x goes to 0, and the integral of 1/x is divergent, so the integral of ln(x+1)/x^2 is divergent.

On (1, inf), we can easily bound ln(x+1)/x^2 by x^n, -2 < n < -1 for all x greater than some value, and the function is integrable below this value, so it is integrable; we do this by bounding ln(x+1) above by x^m, 0 < m < 1 for some m.

For example, let m = 1/2 and x = 9 x^ 1/2 = 3 and ln(9+1) = ln(10) < 3. Also, the derivative of x^(1/2) is

1/2x^(-1/2) and the derivative of ln(x+1) = 1/(x+1) < 1/x, and 1/2 x^(-1/2) > 1/x on (9, inf) iff x^(1/2) > 2 on (9, inf), which is true.

Answer 2

You integrate by parts to get the indefinite integral (ln(x+1) = u and 1/x²dx = dv to get

-ln(1+x)/x + integral of 1/(x(x+1))

The integral can be solved using partial fractions 1/x - 1/(x+1) = 1/(x(x+1))

leading to -ln(1+x)/x + lnx - ln(1+x)

Part a) the integral blows up at the 0 limit and the integral does not converge

Part b) The integral converges as x goes to infinity (limit of 0 at infinity as ln(1+x)/x goes to 0 and ln(x/(x+1)) goes to ln(1) = 0) and you are left with plugging in the lower bound of 1: ln(1+1)/1 - ln(1/1+x) = ln2 - ln(1/2) = ln(4)

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