4
$\begingroup$

Okay, so I am asked to verify the convergence or divergence of the following improper integrals:

$$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$

and

$$\int_1^\infty \frac{x}{1-e^x}dx$$

Now, my first attempt was to use comparison criterion with $$\int \frac{1}{x^2}$$

and conclude that both of the improper integrals converge given that they are smaller than the general term $\frac{1}{x^2}$.

Is it the right path? Also, are the antiderivatives of the improper integrals given easy to find?

Thanks.

$\endgroup$
5
$\begingroup$

For the first one you can write:

$$\int_1^\infty \frac{dx}{1+x^6}\le \int_1^{\infty}\frac{dx}{1+x^2} = \frac{\pi}{2} - \frac{\pi}{4} =\frac{\pi}{4}$$

and

$$\int_0^{1} \frac{dx}{1+x^{6}}\le \int_0^{1} dx=1$$

This shows that

$$\int_0^{\infty}\frac{dx}{1+x^6}$$ exists and is finite. You can similarly show $$\int_{-\infty}^{0}\frac{dx}{1+x^6}$$ exists, and adding the two integrals shows the desired integral converges.

$\endgroup$
3
$\begingroup$

If you exclude some irrelevant finite piece near the origin then $1/x^2$ easily bounds the first integral, yes.

For the second one it's easiest to just use something like $-e^{-x/2}$. The top is eventually smaller than any exponential and the bottom grows exponentially.

Neither has a simple antiderivative, and only the first one has an antiderivative expressible in terms of elementary functions.

$\endgroup$
0
$\begingroup$

For the second, you have two problems. One is at $\infty$, which your comparison will take care of. At $1$ the denominator goes to zero. Once you show it goes to a finite value you have shown convergence.

$\endgroup$
0
$\begingroup$

For the second, note that:

$\int_1^\infty \frac{x}{1-e^x}dx \leq \sum_2^{\infty} \frac{n}{1 - e^{n}}$,

so we can apply the ratio test:

$\lim_{n \to \infty} \frac{(1 -e^n)(n+1)}{(1-e^n)n}$

using L'Hopital once we get:

$\lim_{n \to \infty} \frac{1 - e^n(n+2)}{1 - e^{n+1}(n+1)}$ =

$\lim_{n \to \infty}\frac{1}{e^{n+1}(n+1)-1} + \lim_{n \to \infty}\frac{e^n(n+2)}{1 -e^{n+1}(n+1)}$ =

$\lim_{n \to \infty}\frac{e^n(n+2)}{1 -e^{n+1}(n+1)} = $

(using L'Hopital and simplifying)

$\lim_{n \to \infty} \frac{n+3}{en + 2e} = $

(using L'Hopital again)

$\frac{1}{e} < 1$

so the series, and therefore the integral converges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.