Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Preparing for my calculus exam I found this exercise and I want to see if I'm on the right track

Find out the convergence of the following series and compute it. $$\int_0^\infty \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$

To find if the integral is convergent or divergent I'm thinking to compute this limit: $$\lim_{t\rightarrow\infty} \int_0^t \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$

And if that limit is a constant then my integral is convergent if not then the integral is divergent and I should not compute it.

Now, to compute the integral so I can compute that limit I'm thinking to apply integration by parts.

That $\frac {1}{x+1}$ from the integral is $\frac {d}{\,dx}\ln(x+1)$.

Any tips and/or corrections?

share|cite|improve this question
up vote 4 down vote accepted

Let's try an elementary approach and make the substitution: $$\sqrt{x^2+x+1}=x+t$$ and get $$x=\frac{t^2-1}{1-2t}$$ $$dx=-2 \cdot \frac{t^2-t+1}{(1-2t)^2}\ dt$$ Therefore, our integral gets reduced to $$2\int_{1}^{\frac{1}{2}}\frac{1}{t(t-2)} \ dt=[\log(2-t)-\log(t)]_{1}^{\frac{1}{2}}= \ln 3.$$

Note that for the integration limits, particularly for the case when $x$ tends to $\infty$, I used a celebre limit whose limit is well known, namely:

$$\lim_{x\to\infty} \sqrt[k]{x^k+x^{k-1}+\cdots + 1}-x = \frac{1}{k}, \space k\geq 2$$ that is easily deduced by using Taylor expansion. In our case we dealt with $$\lim_{x\to\infty} \sqrt{x^2+x+1}-x = \frac{1}{2}$$ If you are not used to Taylor expansion, use the substitution $\displaystyle x=\frac{1}{u}$ and pay attention that u tends to $0$. Then you may nicely finish it by using L'Hôpital's rule. Finally, if you don't like either way, just use the conjugates.

Q.E.D. (Chris)

share|cite|improve this answer

If the problem is to examine the convergence alone, just observe that for some constant $C > 0$ we have $$ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq C \min\left\{ 1, \frac{1}{x^2}\right\}. \tag{1}$$ Indeed, on $[0, 1]$, continuity of the integrand shows that it is bounded by some constant $C_1$. On $[1, \infty)$, we have $$ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq \frac{1}{x \sqrt{x^2}} = \frac{1}{x^2}.$$ Thus for $C = \max \{ C_1, 1 \}$ we have $(1)$. Therefore $$ \begin{align*} \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2+x+1}} & \leq C \int_{0}^{\infty} \min\left\{ 1, \frac{1}{x^2}\right\} \; dx \\ & = C \left( \int_{0}^{1} dx + \int_{1}^{\infty} \frac{1}{x^2} \; dx \right) = 2C < \infty. \end{align*} $$

This method is quite general for convergence analysis. When applying this method, we first list all the singularities (the point where the integrand blows up) of the integrand, including points $\pm \infty$ at infinity. For example, if we are dealing with

$$ \int_{0}^{\infty} \frac{dx}{x^{3/2}\sqrt{x + 1}} $$

instead, then our list of singularity will be $\{0, \infty\}$. Then examine the behavior of the integrand near each singularity point $x_0$, by approximating it to familiar functions such as $(x - x_0)^{r}$. In many cases, this information solely determines the convergence behavior of the integral. In our example above, near $x = 0$ the function is approximately $x^{-3/2}$, whose integral near $x = 0$ diverges to infinity. This proves the divergence of the integral above. Rigorous justification of this estimation would be to find a suitable estimation for the integral. In this example, we may argue by

$$ \frac{1}{x^{3/2}\sqrt{x + 1}} \geq \frac{1}{x^{3/2}\sqrt{2}} \quad \text{on} \quad (0, 1]. $$

But in this case, we are asked to find its value. When we succeed in finding its value, then convergence also follows.

We make the substitution $x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan t$. As $x$ ranges from $0$ to $\infty$, $t$ ranges from $\frac{\pi}{6}$ to $\frac{\pi}{2}$. Also differentiating both sides, we have $ dx = \frac{\sqrt{3}}{2} \sec^2 t \; dt$. Thus

$$ \begin{align*} \int \frac{dx}{(x+1)\sqrt{x^2 + x + 1}} &= \int \frac{1}{\left( \frac{1}{2}+\frac{\sqrt{3}}{2} \tan t \right) \left( \frac{\sqrt{3}}{2} \sec t \right)} \cdot \frac{\sqrt{3}}{2} \sec^2 t \; dt \\ &= \int \frac{dt}{\frac{1}{2}\cos t + \frac{\sqrt{3}}{2} \sin t} \\ &= \int \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)}. \end{align*}$$

This shows that

$$ \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2 + x + 1}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)} = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u} $$

for $u = t + \frac{\pi}{6}$. Already it is clear that this improper integral converges, for the integrand is bounded. To find its value, we proceed the calculation.

$$\begin{align*} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u} &= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{\sin^2 u} \; du = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{1 - \cos^2 u} \; du \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ds}{1 - s^2} \qquad (s = \cos u) \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2}\left( \frac{1}{1-s} + \frac{1}{1+s} \right) \; ds \\ &= \frac{1}{2}\left[ \log(1+s) - \log(1-s) \right]_{-\frac{1}{2}}^{\frac{1}{2}} = \log 3. \end{align*}$$

share|cite|improve this answer
With $C=1$. $ $ – Did Sep 1 '12 at 20:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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