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For what values of $\alpha$ is the following integral convergent?


Should the limit comparison theorem be used in this case? I am not familiar with the theorem and would appreciate any help on this.

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Hint: Note that $$ \int_{ - \infty }^\infty {\frac{{|x|^\alpha }}{{(1 + x^2 )^m }}dx} = \int_{ - \infty }^\infty {\frac{{|x|^\alpha }}{{|x|^{2m} }}\frac{{|x|^{2m} }}{{(1 + x^2 )^m }}dx} = \int_{ - \infty }^\infty {|x|^{\alpha - 2m} \bigg(\frac{{x^2 }}{{1 + x^2 }}\bigg)^m dx}. $$

EDIT: For fixed $m$, $(\frac{{x^2 }}{{1 + x^2 }})^m \to 1$ as $|x| \to \infty$. On the other hand, $\frac{1}{{(1 + x^2 )^m }} \to 1$ as $x \to 0$. From these observations one can easily conclude that the integral converges if and only if for any $\varepsilon > 0$ $$ \int_{ 0}^\varepsilon {x^\alpha dx} < \infty $$ (this corresponds, by symmetry of $|x|^{\alpha}$, to the $x \to 0$ case) and for any $M > 0$ $$ \int_M^\infty {x^{\alpha - 2m} dx} < \infty $$ (this corresponds, by symmetry of $|x|^{\alpha-2m}$, to the case $|x| \to \infty$).

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In the second line, we may assume that all integrands are equal to $0$ when $x=0$. – Shai Covo Jun 13 '11 at 14:39

The function is symmetrical about the $y$-axis. Break up the integral into two parts, $-\infty$ to $0$ and $0$ to $\infty$. By symmetry, we need only deal with $$\int_0^\infty \frac{x^\alpha}{(1+x^2)^m}dx$$

We are not told anything about $\alpha$, or indeed about $m$. In particular, $\alpha$ could be negative, which would make the integrand blow up near $x=0$. And of course integrals over an infinite interval raise issues of convergence.

Let's break up the region of integration into two parts, say from $0$ to $1$ and from $1$ to $\infty$. (Instead of $1$ we could use any positive number.)

You probably know that $\int_0^1 x^\alpha dx$ converges if $\alpha>-1$ and diverges if $\alpha \le -1$.

Informally, near $0$, the function $x^\alpha/(1+x^2)^m$ "behaves like" $x^\alpha$, since if $x$ is near $0$, $(1+x^2)^m$ is almost $1$.

So informally, the convergence behaviour of the integral should be the same as the (known) convergence behaviour of the integral of $x^\alpha$.

Semi-formally, we have the following result. Suppose that $f(x)$ and $g(x)$ are positive functions, which are well-behaved everywhere on $(0,b]$ but may blow up as $x \to 0$. If $\lim_{x\to 0}\frac{f(x)}{g(x)}=1$, then $\int_0^bf(x)dx$ converges if and only if $\int_0^b g(x)dx$ converges.

For our integral from $0$ to $1$, let $f(x)$ be our function, and let $g(x)$, the comparison function, be $x^\alpha$. It is very easy to show that the limit of the ratio $f(x)/g(x)$ as $x \to 0$ is $1$. So our integral from $0$ to $1$ converges if and only if $\int_0^1 x^\alpha dx$ converges, and we know everything about that.

Finally we look at the integral from $1$ to infinity. It is useful to deal separately with positive and negative $m$. In fact, from the notation, I expect that implicitly or explicitly $m$ is supposed to be positive.

For the integral to exist, the function must approach $0$ as $x\to \infty$. So we must have $2m>\alpha$. But that's not enough: the function must go down kind of fast.

Informally, if $2m>\alpha$, then for large $x$ our function behaves like $g(x)$, where $g(x)= 1/x^{2m-\alpha}$. (Note that this is not the same $g(x)$ as the one we used for the integral from $0$ to $1$.) We know that $\int_1^\infty 1/x^k dx$ converges if and only if $k>1$.

If our function is $f(x)$, it is easy to show that $\lim_{x\to\infty}f(x)/g(x)=1$.

So we can conclude that our integral from $1$ to $\infty$ converges if and only if $2m-\alpha>1$.

Finally, for our full integral to converge, both parts ($0$ to $1$, $1$ to $\infty$) must converge.

We conclude that our integral converges iff the two conditions $\alpha>-1$ and
$2m-\alpha>1$ both hold.

This can be written more nicely as $$-1 < \alpha <2m-1$$

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Undeleted and edited in response to Theo Buehler's comment.

Observe that the integrand

$$f(x)=\frac{\left\vert x\right\vert ^{\alpha }}{(1+x^{2})^{m}}$$

is even ($f(-x)=f(x)$). We thus have

$$I:=\int_{-\infty }^{\infty }f(x)dx=\int_{-\infty }^{\infty }\frac{\left\vert x\right\vert ^{\alpha }}{(1+x^{2})^{m}}dx=2\int_{0}^{\infty }\frac{x^{\alpha }}{(1+x^{2})^{m}}dx=2I_1+2I_2$$


$$I_1:=\int_{0}^{c}\frac{x^{\alpha}}{(1+x^{2})^{m}}dx,\qquad I_2:=\int_{c}^{\infty }\frac{x^{\alpha}}{(1+x^{2})^{m}}dx,\qquad c>0.$$

This last integral is an improper integral of the first kind with nonnegative integrand. For $x>0$

$$f(x)=\frac{x^{\alpha }}{(1+x^{2})^{m}}=\frac{1}{\left( \frac{1+x^{2}}{% x^{\alpha /m}}\right) ^{m}}=\frac{1}{\left( x^{-\alpha /m}+x^{2-\alpha /m}\right) ^{m}}.$$

Let $g(x)=\dfrac{1}{x^{2m-\alpha }}=\dfrac{1}{\left( x^{2-\alpha /m}\right)^{m}}$. By the limit test

$$\lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty }\left( \frac{x^{2-\alpha /m}}{x^{-\alpha /m}+x^{2-\alpha /m}}\right) ^{m}=\lim_{x\rightarrow \infty }\left( \frac{1}{x^{-2}+1}\right) ^{m}=1$$

and since the standard integral

$$\int_{c}^{\infty }g(x)dx=\int_{c}^{\infty }\frac{1}{x^{2m-\alpha }}dx=\int_{c}^{\infty }\frac{1}{x^{p}}dx$$

is convergent if and only if $p=2m-\alpha >1$, i.e. $\alpha <2m-1$, we conclude that $I_2$ is convergent if and only if $$\alpha < 2m-1.\qquad(1)$$ As for $I_1$ it is an improper integral of the second kind with nonnegative integrand. Let $h(x)=\dfrac{1}{x^{-\alpha}}$. Now the limit test is

$$\lim_{x\rightarrow 0^+}\frac{f(x)}{h(x)}=\lim_{x\rightarrow 0^+}\frac{x^{\alpha }}{(1+x^{2})^{m}}x^{-\alpha}=\lim_{x\rightarrow 0^+}\frac{1}{(1+x^{2})^{m}}=1$$

and since the standard integral

$$\int_{0}^{c }h(x)dx=\int_{0}^{c }\frac{1}{x^{-\alpha}}dx=\int_{0}^{c}\frac{1}{x^{p}}dx$$

is convergent if and only if $p=-\alpha <1$, $I_1$ is convergent if and only if $$-1<\alpha \qquad (2)$$

From $(1)$ and $(2)$ follows that the given integral $I=2I_1+2I_2$ is convergent if and only if $$-1<\alpha<2m-1\qquad (3)$$

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This looks good at a quick glance. I haven't checked all the details, though. Thank you! – t.b. Jun 19 '11 at 16:39
@Theo Buehler: You are welcome! – Américo Tavares Jun 19 '11 at 16:46

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