# Improper integrals with bounds 0 to infinity

$$\int_{0}^{\infty }\frac{1}{x^{1/2}+x^{3/2}}dx$$

I managed to integrate correctly and get $$2\arctan(\sqrt{x})+C$$, but I was wondering why my teacher split it up into two pieces, one from $$0$$ to $$1$$ with a limit at $$0$$ evaluating $$2\arctan(\sqrt{x})+C$$ and the other $$1$$ to $$\infty$$ with a limit at infinity evaluating $$2\arctan(\sqrt{x})+C$$. Can I just take $$2\arctan(\sqrt{x})$$ and evaluate from $$0$$ to $$\infty$$ with a limit at infinity?

• You can do this, but only after you've proven that this doubly-improper (double because of the $\infty$ and the singularity at $0$) integral converges in the standard sense. We define $\int_0^\infty (\cdots)dx$ to be the sum of $\int_0^c (\cdots)dx$ and $\int_c^\infty (\cdots)dx$ provided the latter integrals exist for some $c$; this is why your teacher is splitting the integral into two. Commented Dec 1, 2022 at 1:20
• You can just evaluate your answer from $0$ to $\infty$. Maybe your teacher just wanted to show that evaluating it from $0$ to $1$ gives you $\pi/2$ and evaluating it from $1$ to $\infty$ also gives you $\pi/2$. Commented Dec 1, 2022 at 3:29
• @Accelerator Yes maybe teacher wanted students to know its $\frac \pi2 + \frac \pi2$ but I wonder is it possible to have first sight intuition for the same by looking at the problem only? Commented Dec 1, 2022 at 8:00
• Sure, you can split up the integral like that, but what would be the point? That is just more work. The integrand is smooth everywhere on $(0,\infty)$, so you can just find the indefinite integral and just plug in the limits. That's the intuition everyone else would have. Commented Dec 1, 2022 at 8:12
• @Accelerator Indeed! The integrand is smooth everywhere on $(0, \infty)$ but what fascinated me is that teacher split the integral as if he/she already knew that $\int_0^1f(x)dx = \int_1^\infty f(x)dx = \frac \pi2$ where $f(x) = \frac 1{x^{\frac 12} + x^{\frac 32}}$ Commented Dec 2, 2022 at 5:26

Simpler approach: Let $$y=x^\frac{1}{2}$$. Then $$dy=\frac{1}{2}x^{-\frac{1}{2}}dx$$ or $$dx=2ydy$$. Integral is $$\int_0^\infty 2\frac{1}{1+y^2}dy=\pi$$

My best guess is that it's because we have an improper integral with an issue caused at each boundary point: $$0$$ and $$\infty$$ each cause a certain type of improperness. Working entirely based on $$(0,\infty)$$ can be a bit sketchy, because these improper integrals are defined as limits. Namely: for this type, we can define

$$\int_0^\infty f(x) \, \mathrm{d}x \stackrel{\text{def}}{=} \lim_{a \to 0^+} \int_a^c f(x) \, \mathrm{d} x + \lim_{b \to \infty} \int_c^b f(x) \, \mathrm{d} x$$

for any $$c \in (0,\infty)$$.

Note: You may be thinking of "wait, isn't this just the property of additivity?", where we have $$\int_a^b f(x) \, \mathrm{d} x = \int_a^c f(x) \, \mathrm{d} x + \int_c^b \, \mathrm{d} x$$ for a $$c \in (a,b)$$. While it is motivated by that, this is the definition for this type of improper integral, the starting point.

Opting to work solely over $$(0,\infty)$$ without splitting it up could work, if you're careful. One has to be mindful that (if $$F'=f$$)

$$\int_0^\infty f(x) \, \mathrm{d} x = \lim_{b \to \infty} F(x) - \lim_{a \to 0^+} F(x)$$

which is basically per that definition anyways. Concerns can certainly arise, however, if we let

$$\int_0^\infty f(x) \, \mathrm{d} x = \lim_{t \to \infty} F(t) - F \left( \frac 1 t \right)$$

or any other scheme in which the first summand seems to go to $$F$$ "at infinity" and the second seems to go to $$F$$ at $$0$$. This is useful in the notion of the so-called "Cauchy principal value", but sometimes leads to results for integrals that (under normal definitions) don't converge or exist: $$1/x$$ for instance having an integral of $$0$$ over $$(-1,1)$$.

Put differently, the crux of this latter item being an issue could be thought of as you constraining $$F(t)$$ and $$F(1/t)$$ to go somewhere at the same speed and along a certain path, which could affect the limit, whereas the limit in two variables $$a,b$$ is a bit more free and matches more appropriately the formal definitions of limit.

I bring it up to establish why the initial definition is what we use, contrasted against your own.

Of course, this still may even match the "standard" answer for an integral, as it seems to here, but it's a sort of "right answer for the wrong reasons" sort of deal.

If not wrong: You could solve the problem but you want to know why did your teacher split the above integral as $$\int_0^\infty = \int_0^1 + \int_1^\infty$$ or as $$\arctan(x)|_0^\infty = \arctan(x)|_0^1 + \arctan(x)|_1^\infty$$

however, for this particular case

\begin{align*}\int_0^\infty \frac {dx}{x^{\frac 12} + x^{\frac 32}} &=\left(\int_0^1 + \int_1^\infty\right)\frac {1}{x^{\frac 12} + x^{\frac 32}}dx \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_1^\infty\frac {dx}{x^{\frac 12} + x^{\frac 32}}} ; \text{ Let } \color{green}{x = \frac 1y } \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \int_1^0\frac {\frac{-dy}{y^2}}{y^{\frac {-1}2} + y^{\frac {-3}2}} \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_0^1 \frac {dy}{y^{\frac 12} + y^{\frac {3}2}}} \\& = 2\times\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} \\& = 2\times\int_0^1 \frac {x^{-\frac 12}dx}{1 + x} \\& = 2\frac {\beta\left(\frac 12, \frac 12\right)}{2} = \pi \end{align*}

Why did I use $$\beta$$ function?: Master Theorem