Why $\int_{1}^{\infty} e^{-x} dx$ converges but $\int_{1}^{\infty} \frac{1}{x} dx$ does not?

Question

I'm struggling to intuitively understand the above. They both have similar shape and have an asymptote at y = 0, why is it that the exponential converges but the hyperbola diverges?

Answer 1

Saying "they both have similar shape" is not a reasonable description. It is true that the two functions both decrease towards $0$, but $e^{-x}$ (blue in the first graph below) falls much faster towards $0$ than $\frac1x$ (red) does.

It is visibly plausible that the area below the blue curve extended to the right may be finite while the area below the red curve extended to the right may not be.

If you do the partial integrals then you can confirm this, since $\int\limits_1^y e^{-x} \, dx = e^{-1}-e^{-y}$ and $\int\limits_1^y \frac1x \, dx = \log_e(y)$. Visually you can then see that the blue integral rapidly approaches an upper bound when $y$ increases, while the red integral keeps increasing, albeit at a slightly decreasing rate.

Answer 2

It is a matter of how fast the function tends to zero. Exponential decay is very fast.

$$\int_n^{n+1}e^{-x}dx=e^{-1}\int_{n-1}^ne^{-x}dx$$ and you have a decreasing geometric series.

Harmonic decrease is slow.

$$\int_n^{2n}\frac{dx}x=\int_{n/2}^n\frac{dx}x$$ and you add an infinity of constant terms.

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If you consider

$$\int_1^\infty\frac{dx}{x^a}=\left.\frac{x^{1-a}}{1-a}\right|_1^\infty$$ you understand that a critical exponent is $a=1$. Hence $\dfrac1x$ is slightly too slow to converge.