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Question is to check which of the following improper integrals are convergent?

$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$

$$\int _0^5 \frac{dx}{x^2-5x+6}$$

$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$

I was having a stupid mindset that :

"$\textbf{Improper integrals should have at least one limit infinite}$"

and so i came to conclusion that given two integrals with finite limits are convergent and thus question is nonsense.

But sooner i realized that for two integrals $$\int _0^5 \frac{dx}{x^2-5x+6}$$

$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$

denominators have zeros in given interval $[0,5]$

So, then I decided that these integrals are actually improper :)

Now, I tried seeing $$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$ as limit of $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}$$ with $b\rightarrow \infty$

$$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=2\sqrt{x^2+2x+2}.\frac{1}{2x+2}=\sqrt{x^2+2x+2}.\frac{1}{x+1}=\sqrt{\frac{(x+1)^2+1}{(x+1)^2}}=\sqrt{1+\frac{1}{(x+1)^2}}$$

sorry for forgetting limits

which would imply that $$\int _1^{b} \frac{dx}{\sqrt{x^2+2x+2}}=\sqrt{1+\frac{1}{(b+1)^2}}-\sqrt{1+\frac{1}{(1+1)^2}}$$

and when $b\rightarrow \infty$ as $b$ is in denominator... this limit exists and so,
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}=1-\sqrt{\frac{5}{4}}$$

As $x^2+2x+2$ is positive in $[1,\infty]$, the integral should be positive, I might have missed some signs in between but i am sure that this is convergent.. (I would be thankful if some one can help me to see where the problem is with signs)

Coming to second problem, $$\frac{1}{x^2-5x-6}=\frac{1}{x-3}-\frac{1}{x-2}$$

$$\int _0^5 \frac{dx}{x^2-5x+6}=\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$

$$\int _0^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)=\int _0^2 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _2^3 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)+\int _3^5 \big(\frac{1}{x-3}-\frac{1}{x-2}\big)$$

$$=\log (2-3)- \log(2-2) \dots$$ but then $\log 0$ is not defined

SO, I would like to confirm that this integral is divergent.

Coming to third integral $$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$

I would like to see this as limit of $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$

and $$\int _a^5\frac{dx}{\sqrt[3]{7x+2x^4}}=\frac{3}{2}(7x+2x^4)^{\frac{2}{3}}.\frac{1}{8x^3+7}$$

with similar calculations as i have done for first integral, I have seen that this integral also converges (I do not want to write that and kill readers patience :D)

I am sure if my idea for first integral is correct then third integral should also converge.

So, I would say that first and third integrals converge where as second integral do not converge.

I would be thankful if someone can verify my solution and please let me know if there are any gaps.

Thank You.

P.S : I would like to remind for users (who are as dumb as i am) that

$\textbf{Improper integrals need not necessarily have infinite end points}$

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  • $\begingroup$ Please check your integration rules, e.g. $$\int \frac{dx}{\sqrt{x^2+2x+2}} = \textrm{arcsinh(x+1)} + C$$ and $\int_1^b(...) dx$ is a function of $b$ not $x$. $\endgroup$ Nov 22, 2013 at 12:47
  • $\begingroup$ @gammatester : Yes Yes... for $\int _1^b (...)dx$ I have typed for a long time and so i have asked for an excuse "sorry for forgetting limits".. I understand that i should have explicitly write that.... Thank you for mentioning.. :) $\endgroup$
    – user87543
    Nov 22, 2013 at 15:09

1 Answer 1

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Hint: For the issue concerning convergence it sufficient and necessary to understand only the two following types of integrals for $\alpha \in \mathbb{R}$: $$ \int_{0}^1 x^\alpha d x , \qquad \int_{1}^\infty x^\alpha d x.$$ E.g. $$ \int_{1}^\infty ( (x+1)^2 + 1)^{-1/2} d x = \int_{0}^\infty ( x^2 + 1)^{-1/2} d x \geq C * \int_{1}^\infty x^{-1} dx = \infty$$

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  • $\begingroup$ I am sorry I could not follow your details given... could you please elaborate your idea... $\endgroup$
    – user87543
    Nov 22, 2013 at 15:10
  • $\begingroup$ Do you understand the example? $\endgroup$
    – Marc Palm
    Nov 22, 2013 at 15:14
  • $\begingroup$ yes yes.. now i understood... Thank you $\endgroup$
    – user87543
    Nov 22, 2013 at 15:16

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