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I am at a loss for what to do.

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx$$ I tried to make $u = x^6 + 9 $, $du = 6x^5$

$$\frac{1}{5}\int_{-\infty}^\infty \frac {1}{x^3u}du$$

I can rewrite $x$ in a complex manner but I do not think that actually helps me.

I tried to do some algebra magic be rewriting $x^6$ as $x^3 \cdot x^3$ but I made no real progress like that. I know I can make it $(x^3 + 3)^2 - 6x$ but that doesn't seem to do any good.

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It would be nice to have $x^2$ be more or less the derivative of $u$, so you might want to try $u=x^3$. – Phira Jun 7 '12 at 23:10
@Phira Is there a rule or process to follow to see that? – user138246 Jun 8 '12 at 1:10
@Jordan It should be natural in the hierarchy of your mathematical toolkit you've learned in calc 1 and 2. If you can use u-substitution over integration by parts, partial fractions, trig substitutions, etc, you should. When using u-substitution, you're always looking for the $du$ to be part of integrand already - it makes things easy. – Joe Jun 8 '12 at 2:37
I was simply trying to help you understand the substitution made for your problem, not give you life advice. – Joe Jun 8 '12 at 2:42
Telling me it is something I should already know isn't useful, obvsiously I do not know this. – user138246 Jun 8 '12 at 2:42
up vote 1 down vote accepted

This is supplementary to Peter and Michael's answers (and Phira's comment).

Here is what a plot of $\tan x$ looks like between $x=-\pi/2$ and $x=+\pi/2$:

$\hskip 2in$ tan

We have a couple one-side limits from the above, the first from the right and second from the left:

$$\lim_{x\to-\pi/2^+}\tan x=-\infty,\qquad \lim_{x\to+\pi/2^-}\tan x=+\infty. \tag{$\circ$}$$

This isn't terribly difficult to see visually by drawing right triangles with $\theta\approx\pm\pi/2$. Alternatively, we could find the left/right limits of $\sin$ and $\cos$ separately at $\pm\pi/2$ and reason the above limits.

Symmetrically, here is what $\arctan x$ looks like on the real line:

$\hskip 2in$ arctan

Above is the graphical depiction of why the limits $(\circ)$ are reversible. That is,

$$\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2},\qquad \lim_{x\to+\infty}\arctan x=+\frac{\pi}{2}. \tag{$\bullet$}$$

This is what allows you to finish off the computation method given by the substitution $u=x^3$. In my opinion this is the standard method that would be intended for calculus students, and the limits given in $(\circ),(\bullet)$ were intended to be in your personal "arsenal" a priori for this problem.

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I am still confused on how I am suppose to calculuate, without the use of graphics, the limit, I don't need to graph every function I find the limit of, so how do I do this one? – user138246 Jun 8 '12 at 1:38
Like I said, I think it was intended you know this limit a priori - it was not supposed to be new information for you, but an old fact to bring forth. I don't know if your class or textbook or whatnot actually covered it, though. Note that triangle edge ratios are a reasonable definition of the trigonometric functions, so evaluating the limits by reasoning with them would be reasonable too. Otherwise, as I alluded, one can use $\sin\to\pm1$ and $\cos\to0$ as $x\to\pm\pi/2$ (respectively) to get $(\circ)$, and then $(\circ)\implies(\bullet)$. – anon Jun 8 '12 at 1:51
The last implication, if you really want something formal, can be done with limiting substitution. Namely, if $x\to\pm\infty$, we can replace it with $x=\tan\theta$ as $\theta\to \pm\pi/2^{\mp}$ respectively inside the limits of $(\bullet)$. – anon Jun 8 '12 at 1:54
But I have arctan, that isn't really related in any way to sin or cos is it? – user138246 Jun 8 '12 at 1:57
$\tan=\sin/\cos$ – anon Jun 8 '12 at 2:01

You are given

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx$$

Note first that the function in question is even. For any even function, we have that

$$\int_{-a}^{a}f(x) dx = 2\int_0^a f(x) dx$$. This can be generalized for improper integrals. So we have that

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx=2\int_0^\infty \frac {x^2}{x^6 + 9}dx$$

Now we can make an appropriate change of variables. Note that $x^6 = (x^3)^2$. We let $x^3=u$, then we get $3x^2 dx=du$. This works, because the expression $x^2 dx$ appears in the integral. Now we write the integral in terms of $u$ - remember we have to replace every instance of $x$ in the integral with $u$.

Note that when $x \to \infty$, $u \to \infty$, and $x \to 0$, then $u \to 0$. We get

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx=\frac 2 3\int_0^\infty \frac {du}{u^2 + 9}$$

I belive you can evaluate this integral now.

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I end up with a limit of $1\9 arctan(x^3 /3$ and I have no clue how to evalutate that. – user138246 Jun 7 '12 at 23:31
Can you find $\ell =\lim\limits_{x \to \infty} \arctan x$? – Pedro Tamaroff Jun 7 '12 at 23:34
No I am not sure how to do that. – user138246 Jun 7 '12 at 23:34
Are you aware that $\lim\limits_{x \to \pi/ 2^+}\tan x = +\infty$? – Pedro Tamaroff Jun 7 '12 at 23:36
Well, the integral you have is called an improper integral, so for the $0$ I guess you see that $\arctan 0=0$. Then, for the upper limit, you have to take $x \to +\infty$. As a general result, the $\arctan$ function has the property that $x \to \pm \infty$, then $\arctan x \to \pm \pi/2$. That follows from the way we define the inverse tangent, and how the tangent behaves for $\pi /2^-$ and $-\pi /2^+$. Errata: in my last comment, it should read $x \to \pi/2^-$. Maybe you should re-read your trig notes. – Pedro Tamaroff Jun 7 '12 at 23:41

$$u = x^3, \quad du = 3x^2\,dx$$ Then $$ \int \frac{x^2\,dx}{x^6 + 9} = \int\frac{du/3}{u^2+9}. $$ You get an arctangent.

Be sure to learn this: The expression $x^2\,dx$ begs for letting $u$ be a thrid-degree polynomial, because $x^2$ is the derivative of a third-degree polyonomial.

Later addendum: $$ \int\frac{du/3}{u^2+9} = \int\frac{du/3}{9\left(\frac{u^2}{9}+1\right)} = \frac 1 9 \int \frac{du/3}{(u/3)^2 + 1} = \frac 1 9 \int \frac{dw}{w^2 + 1} $$ $$ = \frac 1 9 \arctan w + C = \frac 1 9 \arctan \frac u 3 + C = \frac 1 9 \arctan \frac{x^3}{3} + C. $$

Recall from trigonometry that $\arctan v\to\pi/2$ as $v\to\infty$ and $\arctan v\to-\pi/2$ as $v\to-\infty$. And $x^3$ approaches $\infty$ as $x\to\infty$ and similarly $x^3\to-\infty$ as $x\to-\infty$. Bottom line: $\pi/9$.

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I am not sure what to do next. – user138246 Jun 7 '12 at 23:29
I am not sure what to do next. I am not sure if this method works, I need to have atleast one number to work with and this just leaves two infinity values which are useless. – user138246 Jun 7 '12 at 23:36
@Jordan $$\begin{eqnarray*} \left. \frac{1}{9}\arctan \frac{x^{3}}{3}\right\vert _{-\infty }^{+\infty } &=&\lim_{x\rightarrow +\infty }\frac{1}{9}\arctan \frac{x^{3}}{3} \\ &&-\lim_{x\rightarrow -\infty }\frac{1}{9}\arctan \frac{x^{3}}{3} \end{eqnarray*}$$ – Américo Tavares Jun 8 '12 at 0:10
@AméricoTavares What about the undefined value and pi/2? It seems like I get an infinity over infinity value. – user138246 Jun 8 '12 at 0:11
@Jordan $\lim_{x\rightarrow +\infty }\frac{1}{9}\arctan \frac{x^{3}}{3}$ and $\lim_{x\rightarrow -\infty }\frac{1}{9}\arctan \frac{x^{3}}{3}$ are finite, they are not undefined. – Américo Tavares Jun 8 '12 at 0:14

This method will evaluate the integral, however not in an basic way using substitution, as I assume you want. I just think some will find this method interesting.

Letting $$f(z)=\frac {z^2}{z^6 + 9}$$

Integrating in the positive sense around the contour formed by a semicircle of radius $R$ (called $C_R$) on the complex plane, we have

$$\int_{C_R}f(z)\, dz=\int_{-R}^R f(z)\, dz+\int_\text{Arc} f(z)\, dz$$

As $R \to \infty$, $\int_\text{Arc} f(z)\, dz=0$, thus

$$\lim_{R \to \infty} \int_{C_R}f(z)\, dz=\int_{-\infty}^\infty f(z)\, dz = 2 \pi i\sum \text{Residues of f(z) in }\lim_{R \to \infty}C_R$$

The poles, $z_1$, $z_2$ and $z_3$ of $f(z)$ are (with de Moivre's theorem)

$$z_1=3^{1/3}\exp(\frac{i \pi}{6})$$ $$z_2=3^{1/3}i$$ $$z_3=3^{1/3}\exp(\frac{5 i \pi}{6})$$

The residues of the poles, $b_1$, $b_2$ and $b_3$ respectively, are (I used L'Hopital's rule and the limit definition of the residue, but an alternative may be possible)

$$b_1=-\frac{i}{18}$$ $$b_2=\frac{i}{18}$$ $$b_3=-\frac{i}{18}$$

So finally,

$$\int_{-\infty}^\infty f(z)\, dz=2\pi i (-\frac{i}{18}+\frac{i}{18}-\frac{i}{18})=-\frac{\pi i^2}{9}=\frac{\pi}{9}$$

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So you need about 6 years of college math to understand this answer? – user138246 Jun 8 '12 at 0:18
@Jordan I'm not in any university, my friend, nor have I ever been. I have only an interest in mathematics. Anyone who wants to learn complex integration will be able to understand my answer. I don't expect you to understand this, I am merely contributing another interesting method to evaluate the integral. – Argon Jun 8 '12 at 0:22
Plus, I object to posting a complete answer to a homework question within one hour! – GEdgar Jun 8 '12 at 0:31

In this situation you can either adopt real analysis methods (as shown very elegantly by the posters prior to me), or you can use the residue theorem, a complex analysis technique.

We have an even integrand of a real definite integral from the negative infinity to positive infinity, define $f(z)=\frac{z^2}{z^6+9}$ and define $C$ to be a semicircle on the upper half of the complex plane with a radius large enough to enclose any singularities of $f$.

We say $\int_C f(z) \, dz = \text{answer} = 2\pi i \operatorname{Res}_f$

Solving $z^6=-9$ gives us 6 imaginary solutions, 3 of which are on the upper plane, they are: $\alpha=9^{1/6}\exp(i \pi /6)$, $\beta=9^{1/6} \, i$, $\gamma=9^{1/6} \exp(5i \pi /6)$

Using the residue theorem, we differentiate the denominator of f and evaluate this new function at alpha, beta and gamma and multiply by $2\pi i$ to find the integral:

We evaluate $\pi i z^{-3}$ at alpha. beta and gamma and sum up our results to obtain $\frac{\pi i}{3} 9^{-1/2} (-i)= \frac{\pi i}{9}$

My answer is rough (written on an iPad), but as this is homework perhaps seeing a different method may help you understand the question!

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I just posted a complex analysis solution here: . You may have missed it. – Argon Jun 8 '12 at 0:34
Ah it is very unfortunate we posted the same solution, I began working on my mine around 20 minutes prior to posting it (using a tablet) and did not refresh the page! Either way, your solution explains everything step by step – Arbias Hashani Jun 8 '12 at 1:13

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