# How come that $\int_{0}^{1}\frac{1}{\sqrt{x}}\mathrm dx$ converges?

I cant understand why this integral converges $\int_{0}^{1}\frac{1}{\sqrt{x}}\mathrm dx$.

If you take a look at the graph of $\frac{1}{\sqrt{x}}$ between $0$ and $1$ it never gets a value for $x = 0$. So I thought that the line continues into infinity without touching the y-axis, so that'd make it that the surface area is infinite. However when solving the integral, I get a real number?

• It is the same principle behind the convergence of an infinite series. Think about $\sum_{n=1}^\infty 1/n^2$: you are constantly adding strictly positive numbers to it, so you would expect the sum to be infinite. Yet it converges to $\pi/6$.
– John
Commented Feb 2, 2017 at 13:07
• @John So If I understand correctly, it is because the numbers you are adding when coming close to 0 are so small, that it doesnt really have an impact on the total value of the integral? Commented Feb 2, 2017 at 13:12
• Exactly! You can actually see it from the definition of integral based on Riemann sums (or Darboux sums)
– John
Commented Feb 2, 2017 at 13:15
• @John a little typo - it sould be $\pi^2/6$. Commented Feb 2, 2017 at 13:19
• @TZakrevskiy thanks!
– John
Commented Feb 2, 2017 at 14:20

Since $\int_{0}^{1}\frac{1}{x^\frac{1}{2}}dx$ with $p=\frac{1}{2}<1$, by the p-Test the integral is convergent.

We can analyse this in the following way:

$$\int_{0}^1\frac{1}{\sqrt{x}}dx$$

$$=\lim_{a\rightarrow 0}\int_{a}^{1}\frac{1}{\sqrt{x}}dx$$

$$=\lim_{a\rightarrow 0}[2\sqrt{x}]_{a}^{1}$$

$$=\lim_{a\rightarrow 0}(2-2\sqrt{a})$$

$$=2.$$

• You miss $dx$ in your notation, and simply working out the integral doesn't explain why an integral over unbounded function converges. Commented Feb 2, 2017 at 13:25
• ^ Edited. Thanks for your feedback. Commented Feb 2, 2017 at 13:34

The line gets infinitely large as we approach $x = 0$, yes, however it also gets infinitely close to the $y$-axis, so the area added, is actually finite. The area under the curve is dependent on both the width and height of the rectangles under, and as such if either dimension grows without bound, the other has the tend to $0$ fast enough as well (for the integral to converge).

There is some tension between the height and the width. Let me explain. Let $0<p<1$, then the integral $$\int_0^1 \frac{1}{x^p}\,\mathrm d x$$ converges. Remember that your example is just with $p = 1/2$, because $x^{1/2} = \sqrt{x}.$ Now if $p \geq 1$, the integral actually diverges. So there is a turning point at $p = 1$, where the width doesn't decrease quite fast enough. This is because for $0<p<1$, we have that $$\int_0^1\frac{1}{x^p}\,\mathrm{d}x = \frac{1}{1-p},$$

from which it is quite easy to see that the value blows up as $p\to 1$.

Consider now the integral $$\int_1^\infty\frac{1}{x^p}\,\mathrm{d}x$$ which converges for exactly $p>1$. In this case, if $0<p\leq1$, then it is the height of the curve as $x\to \infty$ that doesn't go to zero quite fast enough, and hence the integral diverges. For $p>1$ we have that $$\int_1^\infty\frac{1}{x^p}\,\mathrm{d}x = \frac{1}{p-1},$$ and again it is simple to see that this blows up as $p\to 1$.

The interesting thing is that both of these integrals diverge exactly when $p = 1$, and so this is the point where both the height as $x\to \infty$ and the width $x\to 0^+$ does not get small quickly enough. For all values of $0<p<\infty$ except $p = 1$ one of the integrals converges.

You hopefully now that the Riemann-ingegral can be seen as the sum of rectangles becoming smaller more and more. Although the surface is not bounded from below the area between the graph and the y-axis become smaller and smaller in width.

But the area of a rectangle is the product of height AND width so if the width become small enough the area keeps small although the height can have a very high value…

So literally although the approximating rectangles become higher and higher the width decreases fast enough to keep the area small so the sum of alle rectangles still converge…

One way to understand this phenomena is by doing the following change of variable: Let $u = \frac{1}{x}$. Then $du = -\frac{1}{x^2}dx$ and we have $$\int_0^1 \frac{1}{\sqrt{x}}dx = \int_{+ \infty}^1 \sqrt u \bigg(-\frac{1}{u^2}\bigg)du = \int^{+ \infty}_1 \frac{1}{u^{3/2}} du.$$ The integrant on the right $\frac{1}{u^{3/2}}$ is finite in every point of the interval $I = [1, \infty)$. The difference here is the the domain of integration is infinite. However this integral on the right is still finite because of the some similar reasons the sum $\sum_{k=1}^n \frac{1}{n^{3/2}}$ is finite.

• You mean $\sum_{n=1}^\infty$ at the end, not $\sum_{k=1}^n$. (Also, why is it finite?) Commented Feb 2, 2017 at 13:54