# Integrals Converging

There always seems to be a question about whether or not an integral converges. Can I ask what the best mental method is to pick the right test/process to calculate the integral? Let’s take an example. $$\int\limits_{0}^{\infty} \frac{ \sqrt{(x^3+1)}}{(x+1)^3} dx$$

For example, I usually start by looking at the limits and looking at where the problem is. So in the example the problem is at $x= \infty$ .

Then i look at what happens as $x \to \infty$. The top behaves like $(x^3)^{0.5} = x^{1.5}$ and the bottom behaves like $x^3$. So the whole thing behaves like $g(x) = x^{-1.5}$.

We know for all $x \in (0,\infty ): ({x^{3+1}})^{0.5}/(x+1)^3 \leq x^{-1.5}$, so by the comparison test we can say that integral $g(x)$ converges so the original integral does.

Q1. Do you think this answer would get me full marks in an exam? Is anything key missing? Q2. Can you always use the Comparison Test? Q3. When do you use the Limit Comparison Test vs the comparison test?

In the above if we divide f(x) (original integral) by g(x) then we get another weird integral which idk how to solve.

Q4. We can make substitutions, split the integral etc right?

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Similar to Joriki's comments on your other posts, I will start to hold you to much higher standards on this site. I recommend that you start showing your appreciation to the people who have helped you and that you read the FAQ. – mixedmath May 27 '11 at 23:20

Your approach is correct. If you had an improper integral of mixed type, you would have additionally a singularity in one of the limits of integration of the integrand $f(x)$, and you should analyze the convergence or divergence, by comparing $f(x)$ to a function $g(x)$ with a singularity at that limit.

Sometimes to find the function $g(x)$ to be used in the limit comparison test, it is possible to transform the integrand by manipulating it algebraically. In your example, we can make the following calculation:

$$\begin{eqnarray*} f(x) &=&\frac{\sqrt{x^{3}+1}}{(x+1)^{3}}=\frac{\sqrt{x^{3}\left( 1+x^{-3}\right) }}{(x+1)^{3}}=\frac{x^{3/2}\sqrt{1+x^{-3}}}{(x+1)^{3}} \\ &=&\frac{x^{3/2}\sqrt{1+x^{-3}}}{x^{3}\left( 1+3x^{-1}+3x^{-2}+x^{-3}\right) } \\ &=&\frac{1}{x^{3/2}}\cdot \frac{\sqrt{1+x^{-3}}}{1+3x^{-1}+3x^{-2}+x^{-3}}. \end{eqnarray*}$$

Since

$$\lim_{x\rightarrow \infty }\frac{\sqrt{1+x^{-3}}}{1+3x^{-1}+3x^{-2}+x^{-3}}=1,$$

we conclude, as you did, that the integrand behaves like $g(x)=\frac{1}{x^{3/2}}$, the limit being

$$\lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}=1.$$

We know that $\int_{a}^{\infty }g(x)$ $dx$ is convergent for $a>0$, so is

$$\int_{0}^{\infty }f(x)dx=\int_{0}^{a}f(x)dx+\int_{a}^{\infty }f(x)dx.$$

Another technique one can use is to find the Taylor series or asymptotic series of $f(x)$ about the point of infinity

$$f(x)=\frac{\sqrt{x^{3}+1}}{(x+1)^{3}}=x^{-3/2}+O\left( x^{-5/2}\right) .$$

Concerning the comparison test to prove convergence we need to find an upper bound for a positive integrand $f(x)$. When $f(x)$ is of the form $f(x)=N(X)/D(x)$, an upper bound $g(x)=M(x)/C(x)$ is obtained by finding functions $M(x),C(x)$ such that $N(x)\le M(x)$ and $C(x)\le D(x)$.

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nice. does this 'method' work for all? – user4645 May 30 '11 at 17:01
@user4645: Comparatively the algebraic manipulation works better in the case when the integrand is composed of polynomials and/or radicals of polynomials, e.g. $(ax^3+bx^2+cx+d)^{1/5}$. – Américo Tavares May 30 '11 at 20:32

You approached the problem in exactly the right way. First you asked where any potential problems lay. It was clear that there is only an issue with the "infinity" part.

If there are two issues, say at $0$ and at infinity, then you should automatically break up the integral at somewhere convenient, like $17$. Almost never will you be able to fix two issues with one idea.

You then asked yourself the right question, how fast does my function approach $0$? If the function decays fast as $x \to \infty$, the integral converges. And if the function does not change sign, and decays slowly, the integral diverges. Note that the opposite is true when the function blows up at a finite point, like $0$. Then slow blowing up is good, and fast is bad.

You gave the right answer to the question "does it go down fast enough?", and drew the right conclusion. You ask whether you would get full marks. In a first calculus course, particularly one not directed to Math students, perhaps you would. In an analysis course, definitely not, since part of the point of an analysis course is to give precise justification for one's intuitions.

Now let's take a more detailed look at your example. If you are going to use comparison, then you need to justify any non-obvious inequalities that you use. And an inequality being "sort of" right is not good enough.

Look at the top. You cannot simply replace $x^3+1$ by $x^3$, since $x^3$ is smaller than $x^3+1$. You are trying to come up with a "good" function $g(x)$ which is $\ge$ your function.

Thus you want to replace the top by something larger, but not too much larger. Note that it is enough for any inequality you write to be true from some point on.

One thing you could do is to observe that if $x \ge 1$, then $x^3+1 \le x^3+x^3$. Also, the bottom is clearly $>x^3$. So the whole thing is, for $x \ge 1$, less than $2^{1/2}x^{3/2}/x^3$, which is $2^{1/2}/x^{3/2}$. But you know that $$\int_1^\infty \frac{1}{x^{3/2}}dx$$ converges, and therefore so does the same thing decorated with $2^{1/2}$, so your integral (but from $1$ to $\infty$) converges, and the part from $0$ to $1$ gives no trouble. Finished!

Or else you could observe that $x^3+1 \le (x+1)^3$, and therefore your function is $\le 1/(x+1)^{3/2}$. Now if you wish change variables by letting $u=x+1$. Finished!

However, given your initial calculation, it might be easiest to use a limit comparison test, if that is an allowed part of the arsenal. Let $f(x)$ be your function. You decided informally that in the long run $f(x)$ "behaved like" $1/x^{3/2}$. Let's prove that, by evaluating the limit $$\lim_{x \to\infty} \frac{f(x)}{g(x)}$$ Thus you want to evaluate $$\lim_{x \to\infty} \frac{(x^3+1)^{1/2}x^{3/2}}{(x+1)^3}$$

The standard way to do this is to divide top and bottom by $x^3$ in a clever way. At the bottom you should get $(1+1/x)^3$. At the top you should get $(1+1/x^3)^{3/2}$. Now let $x \to\infty$. Finished1

Added: You had asked for general heuristics about what to use when. The comparison test, or limit comparison test, are good in general for well-behaved functions that do not change sign. To apply them, you need a supply of "standard" functions about which everything is known.

As mentioned earlier, be prepared to break up an integral. Limit comparison is often easier than comparison, because most students have insufficient experience handling inequalities.

Your post showed some confusion about the limit comparison process. You complained of getting an integral that is more complicated. But in fact, if you look at the solution above that used limit comparison, you will see that there was no integral involved at all. The limit comparison process merely verified your intuition that for large $x$, the function behaves like $1/x^{3/2}$.

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+1 for your very nice explanation. – Américo Tavares May 28 '11 at 18:00
question - how do i know what (t) to pick? i think thats what im finding really hard. question 2 - use CT when there is a sin/cos etc in the integral and LCT otherwise? – user4645 May 30 '11 at 16:58
@user4645: Do not know what you mean by ($t$). Question $2$: For functions that do not change sign, usually either approach is fine. I guess for integral to infinity of $|\sin(x)|/x^2$, observing that $|\sin(x)| \le 1$ seems more natural. But if want to use CLT, can compare with $1/x^{1.9}$. – André Nicolas May 30 '11 at 18:00