8
$\begingroup$

I need to calculate: $$ \lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}\, dt $$ The result should be $1$.

Is there a quicker way than calculating the primitive function?

I thought about seperating to $\int_0^{x+1} -\int_0^x$ but still can't think of the solution.

$\endgroup$
3
  • $\begingroup$ The searched limit is equal to $1$ $\endgroup$ Jan 31, 2019 at 15:13
  • $\begingroup$ Your chosen title is inappropriate, since there is no improper integral in your question. $\endgroup$ Jan 31, 2019 at 15:27
  • $\begingroup$ Can you find upper and lower bounds of the integrand? $\endgroup$ Jan 31, 2019 at 22:11

6 Answers 6

20
$\begingroup$

Why not apply the MVT? The integral is $$F(x+1)-F(x)=\frac{F(x+1)-F(x)}{x+1-x}=F'(x_0)$$ for an $x_0$ between $x$ and $x+1$. Now $F'(x)$ certainly converges to $1$ if $x$ tends to infinity.

$\endgroup$
15
$\begingroup$

Note that\begin{align}\int_x^{x+1}\frac{t^2+1}{t^2+20t+8}\,\mathrm dt&=\int_x^{x+1}1\,\mathrm dt+\int_x^{x+1}\frac{-20t+7}{t^2+20t+8}\,\mathrm dt\\&=1-20\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt.\end{align}So, all that remains to be proved is that$$\lim_{x\to\infty}\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt=0.$$Can you take it from here?

$\endgroup$
5
  • $\begingroup$ Oh i thought about it too, but wasn't sure how to continue. so i guess i prove it with the squeeze theorem like the answer of mathcounterexamples? $\endgroup$
    – Ido
    Jan 31, 2019 at 15:53
  • $\begingroup$ That's what I would do, yes. $\endgroup$ Jan 31, 2019 at 16:08
  • $\begingroup$ I've got one more question - I now understand the idea of using the squeeze theorem, but is it legitimate to argue that when $x$ goes to $\infty$ than the "t" function goes to $1$, and claim that therefore the result is the limit of $1*(x+1-x)=1$? thanks $\endgroup$
    – Ido
    Jan 31, 2019 at 16:30
  • 1
    $\begingroup$ A proof can be found along these lines, yes. $\endgroup$ Jan 31, 2019 at 17:42
  • $\begingroup$ @Ido: see my answer for a "proof along these lines". $\endgroup$
    – Paramanand Singh
    Feb 1, 2019 at 17:42
12
$\begingroup$

For $t > 8$ you have:

$$0 \le 1 - \frac{t^2+1}{t^2+20t+8} = \frac{20t+7}{t^2+20t+8} \le \frac{20t+8}{t^2} \le \frac{21}{t}$$

Hence integrating those inequalities on $[x,x+1]$:

$$0 \le 1- \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt \le 21 \int_x^{x+1} \frac{dt}{t} \le \frac{21}{x}$$

proving that $\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt =1$.

$\endgroup$
1
  • $\begingroup$ I see, thank you very much :) $\endgroup$
    – Ido
    Jan 31, 2019 at 15:54
3
$\begingroup$

For large values of $x$, the graph of $f(t) = \frac{t^2+1}{t^2+20t+8}$ [you can show this by finding $f'(t)$ and show that $f'(t)$ approaches zero as $t$ approaches to infinity] is approximately horizontal, so the area represented by the integral is approximately a rectangle with width $(x + 1) - x = 1$ and height $f(x) = \frac{x^2+1}{x^2+20x+8}$, which approaches 1 as $x$ approaches infinity.

$\endgroup$
1
  • $\begingroup$ That's a visualisation of the approach via MVT. $\endgroup$ Feb 1, 2019 at 12:09
2
$\begingroup$

This is an easy consequence of the definition of limit.


Note that the integrand $f(t) $ tends to $1$ as $t\to\infty $ and hence corresponding to every $\epsilon >0$ we have a corresponding $M_{\epsilon} >0$ such that $$1-\epsilon <f(t) <1+\epsilon $$ whenever $t>M_\epsilon $. Let $x>M_\epsilon $ and then integrating the above inequality with respect to $t$ in interval $[x, x+1]$ we get $$1-\epsilon <\int_{x} ^{x+1}f(t)\,dt< 1+\epsilon $$ whenever $x>M_\epsilon $ and thus by definition the desired limit is $1$.

There is nothing special about the integrand and its limit and what we have proved above can be summarized as the following

Lemma: Let $f:[a, \infty) \to\mathbb {R} $ be a function which is Riemann integrable on every interval of type $[a, b] $ with $b>a$ and let $f(x) \to L$ as $x\to \infty $. Then $\int_{x} ^{x+1}f(t)\,dt\to L$ as $x\to \infty $.

$\endgroup$
1
$\begingroup$

Use the estimate: $$\frac{t-10}{t+10}<\frac{t^2+1}{t^2+20t+8}<\frac t{t+10}, \ t>1 \Rightarrow \\ \int_x^{x+1} \frac{t-10}{t+10}dt<I(x)<\int_x^{x+1} \frac t{t+10}dt \Rightarrow \\ 1-20\ln \frac{x+11}{x+10}<I(x)<1-10\ln \frac{x+11}{x+10} \Rightarrow \\ \lim_{x\to\infty} \left(1-20\ln \frac{x+11}{x+10}\right)\le \lim_{x\to\infty} I(x) \le \lim_{x\to\infty} \left(1-10\ln \frac{x+11}{x+10}\right) \Rightarrow \\ \lim_{x\to\infty} I(x)=\lim_{x\to\infty} \int_x^{x+1}\frac{t^2+1}{t^2+20t+8}dt=1.$$

$\endgroup$

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.