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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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    Telling me it is something I should already know isn't useful, obvsiously I do not know this.2012-06-08

5 Answers 5

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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 was assuming that there was some sort of formula I could use like finding a derivative or something similar.2012-06-08
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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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    @Jordan Then do re-read your trig notes. The tangent tends to $+\infty$ as $x\to \pi/2^-$. In general, it has vertical asymptotes at any multiple of $\pi /2$.2012-06-08
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$u = x^3, \quad du = 3x^2\,dx$ 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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    ....and more $g$enerally, recalling trigonometry will help with many of the things you've posted questions about.2012-06-08
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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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    Plus, I object to posting a complete answer to a homework question within one hour!2012-06-08
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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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    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 step2012-06-08