# Find convergence of integral and compute it

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?

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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)

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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*}

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With $C=1$.  – Did Sep 1 '12 at 20:47