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

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


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

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?



For the first one you can write:

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


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

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.


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.


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.


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.


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