How to prove that converge $$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}$$ and find this value.
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It indeed converges. My answer is about 1.733941 – Divyansh Garg Aug 27 '14 at 05:30
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It can be written in a closed form as (π - ln(3-2√2))/(2√2) – Divyansh Garg Aug 27 '14 at 05:37
3 Answers
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The answer proves the convergence of the integral.
Here is how you prove the convergence. For large $x$ the integrand behaves as
$$ \frac{\sqrt{x}}{1+x^2} \sim \frac{\sqrt{x}}{x^2}$$
which is an integrable function. Or you can use the inequality
$$ \frac{\sqrt{x}}{1+x^2} \leq \frac{\sqrt{x}}{x^2},\quad x\geq 1. $$
Added:
$$ \int_{1}^{b} x^{-3/2} dx = -2 x^{-1/2}\Big|_{1}^{b}=\dots\, $$
and then take the limit as $b\to \infty$.
Added:
If you are interested in evaluating the integral then start with changing the variables $x=t^2$ which transforms the integral to
$$ I = \int_{1}^{\infty} \frac{2t^2}{1+t^4}dt. $$
Using partial fraction we have
$$I = \int_{1}^{\infty} \frac{1}{t^2+i}dt + \int_{1}^{\infty} \frac{1}{t^2-i}dt $$
To finish the problem you need the result
$$ \int \frac{1}{t^2+a}dt = \frac{1}{\sqrt{a}}\arctan\left(\frac{t}{\sqrt{a}}\right) +C $$
Mhenni Benghorbal
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@RoinerSeguraCubero: You need to evaluate $\int_{1}^{\infty}\frac{1}{x^{3/2}}. $ I think you can do it! – Mhenni Benghorbal Aug 27 '14 at 05:27
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That's an upper bound. This definite integral, or related ones, has been on MSE before. – André Nicolas Aug 27 '14 at 05:33
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@RoinerSeguraCubero: What I gave you is how to prove the convergence? – Mhenni Benghorbal Aug 27 '14 at 05:43
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1@MhenniBenghorbal: You proved the convergence. In addition, the question asked for the value of the (original) integral. – André Nicolas Aug 27 '14 at 05:48
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@AndréNicolas: Yes that's what I did which is correct. Thanks for the comment. – Mhenni Benghorbal Aug 27 '14 at 05:49
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@RoinerSeguraCubero: Are you still intersted in evaluating the integral? – Mhenni Benghorbal Aug 27 '14 at 05:50
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Since the convergence has been established.
For the antiderivative, simplifying the results given by a CAS, the antiderivative is $$\displaystyle\int\frac{\sqrt{x}}{1+x^2}dx=\frac{1}{2 \sqrt{2}}\Big[\log \left(\frac{x+\sqrt{2x} +1}{x-\sqrt{2x}+1}\right)+2 \tan ^{-1}\left(\frac{\sqrt{2x} }{1-x}\right)\Big]$$ and the value of the integral simplifies to $$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}=\frac{\pi +2 \coth ^{-1}\left(\sqrt{2}\right)}{2 \sqrt{2}}=\frac{\pi +\log \left(3+2 \sqrt{2}\right)}{2 \sqrt{2}}$$ I suppose that for the antiderivative has been used $$\displaystyle\frac{1}{1+x^2}=\frac{1}{2}\Big(\frac{1}{1-ix}+\frac{1}{1+ix}\Big)$$ and that later a change of variable $x=t^2$ has been used.
Working a little more the antiderivative,$$\displaystyle\int\frac{\sqrt{x}}{1+x^2}dx=\frac{1}{\sqrt{2}}\Big[{\cot ^{-1}\left(\frac{1-x}{\sqrt{2x} }\right)-\tanh ^{-1}\left(\frac{1+x}{\sqrt{2x} }\right)}\Big]$$
Claude Leibovici
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$$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}$$ $$=\int^{\infty}_0\frac{\sqrt{x}}{1+x^2} - \int^{1}_0\frac{\sqrt{x}}{1+x^2}$$ The first integral can easily be evaluated by contour integration over a large semicircle on the upper half complex plane. $$\int^{\infty}_0\frac{\sqrt{x}}{1+x^2} = \frac{\pi}{\sqrt{2}} $$
The second integral can be evaluated by a detailed procedure as illustrated
Divyansh Garg
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