Convergence of $e^x$ and divergence of $1/x$

Question

When taking the improper integral of $e^x$ we have

$$\int_{-\infty}^{0} e^t dt = 1 \tag{1}$$

So $\lim_{h \to -\infty} e^x = 0$ hence the exponential function converges to zero.

If you try the same thing with $\frac1x$ the improper integral diverges to $+\infty$

$$\int_{1}^{+\infty} \frac{1}{t}dt = +\infty \tag{2}$$

Clearly we see on the graph of $\frac1x$ that it converges to zero, but apparently the convergence is too “slow” compared to the exponential, so the area is still infinite, am I right?

Answer 1

Yes this is the idea, to prove formally the statement we can use that

$$\int_{1}^{\infty} \frac{1}{t}dt=\lim_{a \to \infty}\int_{1}^{a} \frac{1}{t}dt =\lim_{a \to \infty} \log a=\infty$$

or also

$$\int_{1}^{\infty} \frac{1}{t}dt \ge \sum_{k=1}^\infty\frac1{k+1}=\infty$$

Answer 2

Yes, you are right. The speed at which the integrand approaches $0$ is the key. You can see the same phenomenon with series. The harmonic series $\sum (1/n)$ diverges while the geometric series $\sum(1/2^n)$ converges.

Answer 3

More correctly,

$\int_1^x dt/t =\ln(x) \to \infty$ as $x \to \infty$.