Area under a curve

[Darya]

Why does a definite integral from 1 to infinity of 1/x dx doesn't converge, even though a limit of 1/x as x approaches infinity does approach 0? How is then the area under 1/x^2 with the same boundaries equal to 1?

PS Sorry for the lack of notation, hope you get it

 

[topsquark]

This is because they are two different functions. The integral of one does not necessarily have anything to do with the integral of the other. Aside from the integrands both being $$\dfrac{1}{x^n}$$ for different n's there is nothing connecting them. You might as well be asking about $$\int_1^{ \infty } e^{-x} ~ dx$$.

$$\int_{1}^{ \infty } \dfrac{1}{x} ~ dx = \lim_{b \to \infty}\int_1^b \dfrac{1}{x} ~ dx = \left . \lim_{b \to \infty }ln|x| \right | _1 ^b$$ does not converge because $$\lim_{b \to \infty } ln|b|$$ does not exist.

$$\int_{1}^{ \infty } \dfrac{1}{x^2} ~ dx = \lim_{b \to \infty} \int_1^b\dfrac{1}{x^2} ~ dx = - \left . \lim_{b \to \infty } \dfrac{1}{x} \right | _1^b$$ converges because $$\lim_{b \to \infty } \dfrac{1}{b}$$ exists.

If it helps we know that $$\dfrac{1}{x} \geq \dfrac{1}{x^2}$$ on the region of integration. So it is at least possible that even though the integral of 1/x is infinite the integral of 1/x^2 is not.

-Dan

 

[topsquark]

Oops![Oops fixed]

Looks like I didn't put in some integral operators in the above. But you know what I mean.
$$\int_1^{ \infty } \dfrac{1}{x} ~ dx = \lim_{b \to \infty } \int _1^b \dfrac{1}{x} ~ dx$$etc.

-Dan

 

[Darya]

This is because they are two different functions. The integral of one does not necessarily have anything to do with the integral of the other. Aside from the integrands both being $$\dfrac{1}{x^n}$$ for different n's there is nothing connecting them. You might as well be asking about $$\int_1^{ \infty } e^{-x} ~ dx$$.

$$\int_{1}^{ \infty } \dfrac{1}{x} ~ dx = \lim_{b \to \infty}\int_1^b \dfrac{1}{x} ~ dx = \left . \lim_{b \to \infty }ln|x| \right | _1 ^b$$ does not converge because $$\lim_{b \to \infty } ln|b|$$ does not exist.

$$\int_{1}^{ \infty } \dfrac{1}{x^2} ~ dx = \lim_{b \to \infty} \int_1^b\dfrac{1}{x^2} ~ dx = - \left . \lim_{b \to \infty } \dfrac{1}{x} \right | _1^b$$ converges because $$\lim_{b \to \infty } \dfrac{1}{b}$$ exists.

If it helps we know that $$\dfrac{1}{x} \geq \dfrac{1}{x^2}$$ on the region of integration. So it is at least possible that even though the integral of 1/x is infinite the integral of 1/x^2 is not.

-Dan

Thanks! Now it makes more sense : )

 

[Steven G]

Consider 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +....

The last term is going to 0 but the sum is infinity.

Just because the last term goes to zero does not mean that the sum will be finite.

However (for positive terms only) if the last term does not go to 0, then with 100% certainty the sum will go to infinity.