# Convergence of improper unbounded integral

Show that $$\displaystyle \int_{0}^{1} \frac{1}{x^2 + \sqrt{x}} dx$$ converges.

My method : This is a type of improper integral where the function becomes unbounded at the lowe limit of integration and both the limits of integration are finite. I took the given function as $$f(x)$$ being integrated from $$0$$ to $$1$$ and another function $$g(x) = \frac{1}{\sqrt{x}}$$ by taking $$\frac{1}{\sqrt{x}}$$ common from $$f(x)$$ and observing that at $$x=0$$ the convergence and divergence of the function $$f(x)$$ was solely dependent by this $$\frac{1}{\sqrt{x}}$$

I was taught a limit test to find the convergence or the divergence. So it says that when the function is of the unbounded type with finite bounds to integrate :

Find $$\lim_{x \to 0 } \frac{f(x)}{g(x)}$$ which I get as 1 in this question. Since the limit exists and is not equal to zero, whatever is the behaviour of $$g(x)$$, that is the behaviour of $$f(x)$$ also. Therefore by using the test integral I find that $$g(x)$$ diverges therfore $$f(x)$$ should also diverges. But the answer says that it converges. How is this possible ?

Test integral used :

$$\displaystyle \int_{a}^{\infty} \frac{1}{x^p} dx$$ where $$a>1$$; If p $$\leq$$ 1 then it diverges. If p $$>$$ 1 than it converges.

I know that in the test integral $$a>1$$ is given but then my prof used it irrespective of it in other examples of the same type (that is in sums of unbounded type where the limits of integration were $$0$$ to $$3$$ and $$0$$ to $$4\pi$$ in the two examples he gave) . And I also do not understand the $$\infty$$ limit in the integration of test integral as in the given question I need to integrate only from $$0$$ to $$1$$

• This may not be how you want to approach this problem, but you can compute that directly. You end up with a few logs and an arctangent - it's a bit of work but not terribly difficult. Jan 15, 2021 at 16:47
• @DMcMor But at the lower limit of 0 the functions do not exist therefore they shouldn't be integrable according to the theory I have been taught. Jan 15, 2021 at 16:50
• You can still find the antiderivative of $\frac{1}{x^2 + \sqrt{x}}$, say $F(x) = \int \frac{1}{x^2 + \sqrt{x}}\,dx$ and compute $F(1) - \lim_{x\to 0^{+}}F(x)$ to check convergence. Jan 15, 2021 at 17:05

You use the wrong test integral. Instead, you should use $$\int_0^1\frac{1}{x^p}\,dx$$ for the cases $$0 and $$p>1$$.

Notes.

You want to analyze the integral $$\displaystyle\int_0^1\frac{1}{x^2+\sqrt{x}}dx$$, which is fundamental different from integrals of the form $$\displaystyle \int_1^\infty f(x)\, dx$$. These are two different types of improper integrals.

In general, if $$0\le f(x)\le g(x)$$ and $$\int_0^1g(x)dx$$ is convergent, then $$\int_0^1f(x)dx$$ is also convergent.

In your example, $$\displaystyle g(x)=\frac{1}{x^{1/2}}$$.

• I haven't been taught this. Could also add the cases as to when it is convergent and when it is diverging ? Im guessing at p=1 it diverges as log zero becomes infinite. But you havent included 1 in the range of p Jan 15, 2021 at 16:51
• @ShauryaGoyal: Yes, you have parallel cases. The case when $p=1$ is also divergent.
– user9464
Jan 15, 2021 at 16:57

The integral of $$g(x)$$ is $$\displaystyle \int_0^1 \frac{1}{\sqrt{x}}\,dx = 2\sqrt{x}\Big\rvert_0^1 = 2$$, doesn't diverge.

Hint

$$\frac{1}{x^2+\sqrt x}\leq \frac{1}{\sqrt x}.$$

• I did try this only to realize that at the lower limit of 0 both of these functions do not exist. Thus they're not proper and not integrable. Jan 15, 2021 at 16:49
• @ShauryaGoyal: I don't understand your comment. My hint allows you to prove that your integral converges.
– Surb
Jan 15, 2021 at 20:25

We have that as $$x\to 0$$

$$\frac{1}{x^2+\sqrt{x}} \sim \frac{1}{\sqrt{x}}$$

and since $$\int_0^1 \frac{dx}{\sqrt{x}}$$ converges also the given integral converges by limit comparison test.