Does $\int_{0}^{\infty} \frac{1}{x} dx$ Converge?

Question

I'm sure countless people has tried to prove this but this is how I came across this, although I do remember seeing this in the past.

Background

https://www.youtube.com/watch?v=vQ0himyDR2E

I was looking at the proof for Gabriel's horn like many people I disagreed with it instantly as I saw that the volume = cross-section at $x = 1$ and how the answer could be that if integration is simply adding up the cross-section together. However I didn't know each cross-section is multiplied by a $\delta{x}$ so that cross-section volume is negligible at that point.

https://www.youtube.com/watch?v=QLHJl2_aM5Q

I watch this to recap integration and found even simpler formula for the surface area $2\pi r = 2\pi \frac{1}{x}$ as all your really doing is summing the circumference of the cross-section. So based on this proof if you sum up the cross-section multiplied by $\delta{x}$ you get pi, however if you sum up the circumference multiplied by $\delta{x}$ you get $\infty$ (should be 2$\pi$ using same logic) both radius goes from 1 to zero so it should not be the case that one converges whilst other diverges. It seems when you are integrating $\frac{1}{x}$ we somehow proved it's $ln|x|$. I plotted the two graphs and saw the gradient are totally different and also the fact $\frac{1}{x}$ has two asymptotes where by $ln|x|$ only has 1. It does not seems like it's integral especially compared to other reciprocal functions and it's integral (Haven't got a formal proof yet) also an exponential and reciprocal functions are very different. Some people have shown it matches because the series:

$$1/2+1/3+1/4+... = \infty$$

diverges but they forgot to multiply by $\delta{x}$ which would not give the same result.

Attempt 1

To compare the function to the area I had two function $f(x) = e^{x+1}$ and $g(x) = \frac{1}{x}$. Integrating $f$(x) from $-\infty$ to -1 area of it and to see how it differs from $g(x)$.

$$\int_{-\infty}^{-1} -e^{x+1}dx = -e^{0} = -1$$

When $g(x)$ is transformed in this way easier to align the area with $g(x)$ but I am not sure how I could prove or disprove if the area fits perfectly inside $g(x)$?

The integral of $\frac{1}{x}$ must also have the following:

$$\int_0^{\infty} \frac{1}{x}dx = T = 2N + 1$$ $$\int_0^{1} \frac{1}{x}dx = T - N = 1 + \int_1^{\infty} \frac{1}{x}$$ $$T - \int_1^{\infty} \frac{1}{x}dx - \int_1^{\infty} \frac{1}{x}dx = 1$$

This is as far as I got anyone know any further steps that can be taken?

Attempt 2

Thanks for the answers and comments I have updated my title as there is no way it can be 1 if the sum of rectangles under the curve from $1 \le x \le 4$ is larger than 1

In my next attempt I will see if $\ln|x|$ matches the properties of the graph. The properties required before were infinite cases and more or less true with $\infty$ (exception made for $\ln|0|$)

The following formula is to calculate the area when $a \lt 1$ and $1 \lt b \lt \infty$ but using the symmetrical properties of the graph by calculating the:

• 1: area from 1 to b adding
• 1.1: (the area from 1 to $\frac{1}{a}$ subtract rectangle under the curve less then a
• 1.2: this gives the top part of the com a to 1 but you still need to add the rectangle underneath it
• 2: add all of the areas and simplify for simplicity and this should be the same as the direct area from a to b

$$ \int_a^b \frac{1}{x}dx = \int_1^b \frac{1}{x}dx + \biggl[\int_1^{\frac{1}{a}} \frac{1}{x}dx - a(\frac{1}{a} -1)\biggr] + (1 - a) = \int_1^b \frac{1}{x}dx + \int_1^{\frac{1}{a}} \frac{1}{x}dx $$ from what we currently have: $$ \int_a^b \frac{1}{x}dx = \ln|b| - \ln|a| $$ from my calculations: $$ \int_a^b \frac{1}{x}dx = \int_1^b \frac{1}{x}dx + \int_1^{\frac{1}{a}} \frac{1}{x}dx = \ln|b| + \ln|\frac{1}{a}| $$

The calculation and the current function does match $\ln|x|$ and that diverges therefore so does this.

Answer 1

It's not an answer because i can't prove something false.

1) For $x>0$,$\dfrac{\partial}{\partial x}\ln x=\dfrac{1}{x}$

$\displaystyle \lim_{a\rightarrow 0,b\rightarrow +\infty}\int_a^b \dfrac{1}{x}\,dx=\lim_{a\rightarrow 0,b\rightarrow +\infty} \ln(b)-\ln(a)=+\infty$

2) for $x>0,f(x)=\dfrac{1}{x}>0$, $f$ is a decreasing function.

\begin{align}\int_1^{n}\dfrac{1}{x}\,dx=\sum_{k=1}^{n-1}\int_k^{k+1}\dfrac{1}{x}\,dx \end{align} But, for $x\in [k,k+1]$, $1\leq k<n$, $\dfrac{1}{x}>\dfrac{1}{k+1}$ therefore,

\begin{align}\int_1^{n}\dfrac{1}{x}\,dx\geq \sum_{k=1}^{n-1} \dfrac{1}{k+1}\end{align}

But, \begin{align}\displaystyle \lim_{n\rightarrow +\infty}\sum_{k=1}^{n-1} \dfrac{1}{k+1}=+\infty\end{align}

Indeed,

Let, $\displaystyle S_n=\sum_{k=1}^{n-1} \dfrac{1}{k+1}$

\begin{align}S_{2n}-S_n+\dfrac{1}{n}&=\sum_{k=n-1}^{2n-1}\dfrac{1}{k+1}\\ &=\dfrac{1}{n+0}+\dfrac{1}{n+1}+...+\dfrac{1}{n+n-1}\\ \end{align}

There are $n$ terms in the sum and each of them is greater or equal to the last one ($f$ is decreasing), that is $\dfrac{1}{2n}$ therefore,

For $n\geq 1$, \begin{align}S_{2n}-S_n+\dfrac{1}{n}&\geq n\times \dfrac{1}{2n}\\ &\geq \dfrac{1}{2} \end{align}

If $\lim_{n\rightarrow\infty} S_n=L$, a real, was existing one would have: $\lim_{n \rightarrow \infty} \left(S_{2n}-S_n+\dfrac{1}{n}\right)=0$

But it's impossible since for $n\geq 1,S_{2n}-S_n+\dfrac{1}{n}\geq \dfrac{1}{2}$

Therefore, no such $L$ does exist but all terms are positive therefore $S_n$ diverges to infinity.

NB:

Since for $x>0,\dfrac{1}{x}>0$, the value of $\displaystyle \int_0^1 \dfrac{1}{x}\,dx$ is finite, or $+\infty$ but since $\displaystyle \int_1^\infty \dfrac{1}{x}\,dx=+\infty$, $\displaystyle\int_0^\infty \dfrac{1}{x}\,dx$ cannot be a finite value.

Answer 2

Mathematics is full of sums and integrals that "are infinite", and methods to "make them finite". This causes statements like $\sum_{n=1}^\infty n=-\frac{1}{12}$ to not so much "not make sense" as "make sense if you say which method you're using" (zeta function regularization will do in that case). Unfortunately, this integral looks as intractable as the series $\sum_{n=1}^\infty\frac{1}{n}$, which I think remains infinite no matter how much you learn about summability methods.

The substitution $x=u^n$ with any $n>0$ gives $I:=\int_0^\infty\frac{dx}{x}=\int_0^\infty\frac{ndu}{u}=nI$, so $I$ is either divergent or $0$. (Unless, of course, you say every positive value is equally valid by varying $n$, which is good as saying you don't have an integrability method to handle this problem at all.) Every mathematician will insist on the former unless a very specific definition of "integration" justifies the latter in context. Indeed, the facts $$n>0\implies\int_0^1x^{n-1}dx=\frac{1}{n},\,n<0\implies\int_1^\infty x^{n-1}dx=-\frac{1}{n}$$are sometimes used for the regularization $\int_0^\infty x^{n-1}dx=0$. I've definitely seen that somewhere, but I can't find a technical name for it.

Answer 3

FYI, it's hard to understand what exactly you're asking about/for here. But for the titular question:

The integral, as stated in the title, does not converge, or if you want (i.e. using the extended real line):

$$\int_{0}^{\infty} \frac{1}{x}\ dx = +\infty$$

So it cannot be proven to converge, as it does not actually converge at all: any attempt you make to prove the opposite will contain a fallacy (invalid reasoning).

This integral, though, is not the one that represents the surface of Gabriel's horn: geometrically, this integral represents the area between entire right-hand graph of $y = \frac{1}{x}$ and the $x$-axis. Moreover, Gabriel's horn starts when $x$ is $1$, not when it is $0$. Otherwise, the volume will be infinite as well as it then acquires an infinite "flange" due to the asymptote at $x = 0$. Nonetheless, even the integral

$$\int_{1}^{\infty} \frac{1}{x}\ dx$$

diverges. The intuitive reason behind this is that $y = \frac{1}{x}$ doesn't "hug" the $x$-axis "fast" enough, as $x$ increases, to leave a finite area. This is, actually, also what is going on with Gabriel's horn, but it's not this integral directly that you want to look at.

If you want to know how to prove it diverges, you basically can do so by the following method: show that the area can be broken up into an infinite number of areas of a minimum finite size, then add them up: adding up an infinite number of any fixed finite number gives infinity. And here's how.

As you seem to be aware, we have that the indefinite integral is

$$\int \frac{1}{x}\ dx = \ln(x) + C,\ x > 0$$

(we won't worry about the $|x|$ bit - there's actually some subtleties there it'd be best to leave off for now, hence why I just threw a stip that this should only be taken as holding for $x > 0$)

From that, of course, you know that

$$\int_{a}^{b} \frac{1}{x}\ dx = \ln(b) - \ln(a) = \ln\left(\frac{b}{a}\right)$$

i.e. the area from $a$ to $b$. Now here's the trick: Let us choose some finite amount of area, say $A$, that this segment should equal. Heck, for ease, we will just take $A = 1$. That is, we will be seeking to break up the area underneath $\frac{1}{x}$ into area-$1$ slices. If

$$\int_{a}^{b} \frac{1}{x}\ dx = \ln\left(\frac{b}{a}\right) = 1$$

then that means $\frac{b}{a} = e \approx 2.718$, by simple use of exponentials. Thus, if we have a bunch of intervals of integration $[a_1, a_2]$, $[a_2, a_3]$, ... such that

$$\frac{a_{j+1}}{a_j} = e$$

then the area over each one is

$$\int_{a_j}^{a_{j+1}} \frac{1}{x}\ dx = \ln(e) = 1$$

Thus, consider the intervals $[1, e]$, $[e, e^2]$, $[e^2, e^3]$, ... giving segments of exponentially escalating length: each interval is $e$ times wider than its predecessor, and directly adjacent thereto. These intervals tile the entire $x$-axis above $1$ and so the areas above them segment up and exhaust the whole area under $\frac{1}{x}$. Moreover, and this is crucial, there are an infinite number of them. Each such interval has area $1$ over it, and now the whole area is, by adding up,

$$ \begin{align} \int_{1}^{\infty} \frac{1}{x}\ dx &= \left(\int_{1}^{e} \frac{1}{x}\ dx\right) + \left(\int_{e}^{e^2} \frac{1}{x}\ dx\right) + \left(\int_{e^2}^{e^3} \frac{1}{x}\ dx\right) + \cdots\\ &= 1 + 1 + 1 + \cdots\\ &= +\infty \end{align}$$

QED.

Answer 4

Quite simply,

$$\int_{0}^{\infty}\frac{1}{x}dx = \int_{0}^{1}\frac{1}{x}dx + \int_{1}^{\infty}\frac{1}{x}dx$$

$$ = \lim_{a \to 0^{+}}\int_{a}^{1}\frac{1}{x}dx + \lim_{b \to \infty} \int_{1}^{b}\frac{1}{x}dx = \lim_{a \to 0^{+}}( \ln1 - \ln a) + \lim_{b \to \infty}(\ln b - \ln 1)$$

$$ \lim_{a \to 0^{+}}\ln \frac{1}{a} + \lim_{b \to \infty} \ln b$$

$$ = \infty + \infty = \infty$$

Answer 5

As others pointed out, the integral diverges. Its regularized value is $\gamma$, the Euler-Mascheroni constant.

Let us break down the integral into two parts: $\int_0^\infty \frac1x dx=\int_0^1 \frac1x dx +\int_1^\infty \frac1x dx$.

Now, using the following transformation: $\int_0^\infty f(x)\,dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) \, dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)\,dx$ involving Laplace transform we can convert each part into equivalent integrals. For the first one we get the following set:

$\int_1^\infty\frac{1}{x}dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=\int_0^\infty-e^x \text{Ei}(-x)dx=\int_0^\infty-\frac{x-x\ln x-1}{(x-1)^2 x}dx$

For the second one we get:

$\int_1^\infty\frac{1}{x}dx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$

We can pick one integral from the first set, (let it be $\int_0^\infty\frac{1-e^{-x}}{x}dx$) and one integral from the second set (let it be $\int_0^\infty\frac{dx}{x+1}$) and find their difference: $\int_0^\infty\frac{1-e^{-x}}{x}dx-\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\left(\frac{1-e^{-x}}{x}-\frac{1}{x+1}\right)dx.$ This difference is equal to Euler-Mascheroni constant $\gamma$, as can be seen from the plot below:

So, $\int_0^1 \frac 1xdx=\int_1^\infty \frac1x dx+\gamma$ and the whole integral is $\int_0^\infty \frac1x dx=2\int_1^\infty \frac1x dx+\gamma$.

We also know that the harmonic series $\sum_1^\infty \frac1x dx$ has regularized value $\gamma$ and the following identity is known: $\sum_{k=1}^\infty \frac1k-\int_1^\infty \frac1x dx=\int_1^\infty\left(\frac1{\lfloor x\rfloor}-\frac1x\right)\,dx=\gamma.$

So, $\sum_{k=1}^\infty \frac1k =\int_0^1 \frac1x dx$, which is not surprising because this series is the Riemann sum for the integral.

Below are the plots of $\int_0^\infty\frac{1-e^{-x}}{x}dx$ (blue), $\int_0^\infty\frac{dx}{x+1}$ (orange) and $\int_0^\infty\frac1{\lfloor x\rfloor}dx$ (green). Tthe last one is equal to the series, shifted left by 1/2:

The areas between green and orange and blue and orange lines both are equal to $\gamma$. Since $\sum_1^\infty \frac1x dx$ has regularized value $\gamma$, we can conclude that $\int_1^\infty \frac1x dx$ has regularized value zero, and so our whole integral $\int_0^\infty\frac1x dx=2\int_1^\infty \frac1x dx+\gamma$ has regularized value $\gamma$.