3
$\begingroup$

$$ H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y} $$

Let $x = \frac{y - 1}{n - 1}$, or $y = (n-1)x + 1$. Then, $dy = (n - 1) dx$.

$$ H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_0^1 \frac{n - 1}{(n - 1) x + 1} dx = \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx $$

$$ \gamma = \lim_{n \rightarrow \infty} \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\lim_{n \rightarrow \infty} \frac{1 - x^n}{1 - x} - \lim_{n \rightarrow \infty} \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\frac{1}{1 - x} - \frac{1}{x}) dx $$

(The first limit relies on the fact that $0 < x < 1$, so that $\lim_{n \rightarrow \infty} x^n = 0$. The second is solved via. L'Hospital's rule.)

Let $t = 1 - x$. Then, $dt = -dx$.

$$ \gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x} = - \int_1^0 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = \int_0^1 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = 0 $$

Evidently something's wrong with this, but I genuinely can't find it. The only part I find even fishy is how the limit is split in two. Any ideas?

$\endgroup$
  • 8
    $\begingroup$ You interchanged a limit and an integral. That's a good place to look, usually. $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '15 at 21:40
4
$\begingroup$

You obviously took a wrong turn along the way to conclude

$$\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x},$$

since each integral is improper and divergent. The limit/integral switch cannot be justified by uniform convergence, monotone convergence, etc.

As an example where the switch is valid, change variables in the first integral with $x = (1-y/n)$ to get

$$H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y} \\= \int_0^n \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n \frac{dy}{y}\\=\int_0^1 \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n \left(1- \frac{y}{n}\right)^n\frac{dy}{y}.$$

Now you can justify a switch of limit and integral using LDCT to obtain

$$\gamma = \lim_{n \to \infty}(H_n - \ln n)= \int_0^1 \left(1 - e^{-y}\right)\frac{dy}{y}-\int_1^\infty e^{-y}\frac{dy}{y},$$

where the improper integrals converge.

Further manipulation leads to the well-known result

$$\gamma = \int_0^1 \left( \frac{1}{1-x}+ \frac{1}{\ln x}\right) \, dx.$$

Here the improper integral converges, as the singular behavior of $1/(1-x)$ is offset by that of $1/\ln x$ near $x=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.