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While going through an Indian text on Analysis I found a test for convergence of improper integral.It was stated without proof.I tried to prove it..then some doubts pop up...

Statement is this :Let $f(x)$ be bounded and integrable in every closed subinterval of $ (a,\infty)$ where $a >0$. Let $\mu$ be a positive number such that $limit_{x \rightarrow \infty} x^{\mu}f(x)$ exists. If $\mu > 1 $ $\int_{a}^{\infty} f(x)dx$ converges. If $ \mu \leq 1 $ $\int_{a}^{\infty} f(x)dx$ diverges.

My proof :

$limit_{x \rightarrow \infty} x^{\mu}f(x) =L $ then for suitable $\epsilon >0$ we will get $x_{0} $ such that $(-L+\epsilon)< |x|^{\mu}|f(x)| < (L- \epsilon)$ for $x>x_{0}$.

This will lead to to $|f(x)| < (L- \epsilon)|x|^{-\mu}$ then using comparison test we will get as $\mu$>1 $\int_{a}^{\infty} f(x)dx$ converges absolutely.

But this proof cannot be used for discussing divergence.Even though we use left inequality $|x|^{-\mu}(-L+\epsilon)< |f(x)|$ .

There are two problems I felt .One is comparison test is applicable for positive functions. $|x|^{-\mu}(-L+\epsilon)$ need not be positive.Can we solve this problem by taking out $(-L+\epsilon)$ ?

Second problem is this:Even though we got for $\mu \leq 1$ this integral diverge by comparison test we get $\int_{a}^{\infty} |f(x)|dx$ diverges.It does not lead to the divergence of improper integral as stated by theorem.

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When $L \neq 0$, this limit comparison can be use to prove both absolute convergence when $\mu > 1$ and absolute divergence when $\mu \leqslant 1$.

Suppose $\lim_{x \to \infty} x^\mu f(x) = L$. Then for $\epsilon = |L|/2$ there exists $x_0 > 0$ such that when $x \geqslant x_0> 0$ we have

$$| |x^\mu f(x)| - |L|| \leqslant |x^\mu f(x) - L| \leqslant |L|/2.$$

Whence,

$$-|L|/2 \leqslant |x^\mu f(x)| - |L| \leqslant |L|/2\\ \implies |L|/2 \leqslant |x^\mu f(x)| \leqslant 3|L|/2 \\ \implies (|L|/2)x^{-\mu} \leqslant | f(x)| \leqslant (3|L|/2)x^{-\mu} $$

and the integrals of $|f(x)|$ and $x^{-\mu}$ must converge or diverge together.

The limit comparison test is often presented only for the case where $f(x) \geqslant 0$ for all $x > a$. In that case, if $\lim_{x \to \infty}x^\mu f(x) = L > 0$, then there exists $x_0$ such that when $x \geqslant x_0$ we have

$$0 < \frac{L}{2}x^{-\mu} < f(x) < \frac{3L}{2}x^{-\mu},$$

and $\displaystyle \int_a^\infty f(x) \, dx$ diverges if $\mu \leqslant 1$ (using the left inequality) and converges if if $\mu > 1$ (using the right inequality).

An example of the former case is $f(x) = \sin(1/x)$ - where $f$ is eventually positive. Since $\lim_{x \to \infty}x \sin(1/x) = 1$, then $\displaystyle \int_1^\infty \sin(1/x) \, dx$ diverges.

If $L = 0$, then for any $\epsilon > 0$ there exists $x_0$ such that when $x \geqslant x_0$ we have

$$-\epsilon x^{-\mu} < f(x) < \epsilon x^{-\mu},$$

and we can only conclude that $\displaystyle \int_a^\infty f(x) \, dx$ converges if $\mu > 1$ (using the right inequality). An example is the gamma function

$$\displaystyle \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \, dx,$$

where proof of convergence is facilitated using the test function $x^{-2}$.

If $L = 0$, then we cannot conclude divergence if $\mu \leqslant 1$ since the left lower bound is negative and the integral could converge to a positive value.

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  • $\begingroup$ So we cannot prove divergence from this theorem because If $\mu \leq 1$ left side diverge it will lead to the conclusion that integral of $|f(x)| $ do not converge.But it does not lead to the conclusion that integral of $f(x)$ diverge.Am I right..should we change the statement of theorem..? $\endgroup$ – Madhu Mar 2 '15 at 2:03
  • $\begingroup$ @Madhu: I added to the answer. Now I think you are concerned about proving divergence for a function which changes sign infinitely often and where the limit is $0$. In that case this test will not help. $\endgroup$ – RRL Mar 2 '15 at 3:28
  • $\begingroup$ okey now it is solved the statement has to modify slightly...$f(x) \geq 0$ and $L \neq 0$. $\endgroup$ – Madhu Mar 2 '15 at 16:18
  • $\begingroup$ Can you suggest any good book which discuss theorems and proofs of different types of improper integrals in university level... $\endgroup$ – Madhu Mar 2 '15 at 16:20
  • $\begingroup$ Yes, with those restrictions it is true. The usual analysis books (Rudin, Apostol, Bartle, etc.) cover improper integrals to some extent. Bartle is more extensive than the others. A Second Course in Mathematical Analysis by Burkill and Burkill gives more intricate examples. $\endgroup$ – RRL Mar 2 '15 at 16:32

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