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I am trying to find the direct expression of the function $f(t)$ given by $$f(t)= \int_{1}^\infty \frac{\arctan (tx)}{x^2\sqrt{x^2-1}}dx$$ It's hard for me to calculate the integration directly.Should I try the method of interchanging $\frac{d}{dx}$ with$\int$ or$\int$ with $\int$ ? Thanks for help.

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What does it mean to exchange $\int$ with $\int$? – Andrew Uzzell Dec 5 '12 at 9:29

Here is an answer obtained by maple

$$ -\frac{\pi}{2} \, \left( -1-t+\sqrt {1+{t}^{2}} \right) . $$

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Exams become easier if the maple was allowed.Thank you all the same. – C Weid Dec 5 '12 at 9:19
@CWeid: You are welcome. – Mhenni Benghorbal Dec 5 '12 at 10:01
@CWeid: By the way, where did this problem come from? – Mhenni Benghorbal Dec 5 '12 at 10:03
Entrance exam in Chern Institute of Mathematics,2009 – C Weid Dec 5 '12 at 14:24

Presumably this was intended to be done with pen and paper, so here is what they might have had in mind. First note that convergence is excellent, so we may differentiate with respect to $t$ to obtain $$ f'(t) = \int_1^\infty \frac{x}{1+t^2x^2} \frac{1}{x^2\sqrt{x^2-1}} dx.$$ Now let $x^2-1 = u^2$ so that $x\, dx = u\, du$ and $$ f'(t) = \int_0^\infty \frac{u}{1+t^2(u^2+1)} \frac{1}{(u^2+1)\sqrt{u^2}} du = \int_0^\infty \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1} du.$$ Putting $$ g(u) = \frac{1}{2} \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1} \quad \text{we seek to evaluate} \quad \int_{-\infty}^\infty g(u) du.$$ This may be done with the Cauchy Residue theorem applied to a half-circle contour in the upper half plane of radius $R$ (the contribution of the circle vanishes in the limit), giving $$ 2 \pi i \left(\operatorname{Res}_{u=i} g(u) + \operatorname{Res}_{u=\sqrt{-\frac{1}{t^2}-1}} g(u) \right).$$ These poles are both simple and we obtain $$ 2\pi i \left(- \frac{1}{4} i - \frac{1}{4} \frac{1}{\sqrt{-\frac{1}{t^2}-1}}\right) = \frac{\pi}{2} - i \frac{\pi}{2}\frac{1}{\sqrt{-\frac{1}{t^2}-1}} = \frac{\pi}{2} - \frac{\pi}{2}\frac{t}{\sqrt{t^2+1}} \quad( t\geqslant 0)$$ Integrating we have $$ f(t) = C + \frac{\pi}{2} t - \frac{\pi}{2} \sqrt{t^2+1}.$$ Finally, to determine $C$, note that $f(0) = 0$ or $$ 0 = C - \frac{\pi}{2} $$ and hence $C = \frac{\pi}{2}$ giving the result $$ f(t) = \frac{\pi}{2} ( 1 + t - \sqrt{t^2+1} ).$$ When $t<0,$same calculation shows $$ f(t)= \frac{\pi}{2}(-1 + t +\sqrt{t^2+1})$$

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To differentiate the function under the integral,we need to check the convergence of $f'(t)$,whether it's uniform,rather than $f(x)$ itself.Thank you for this excellent answer,I am trying to find a simpler one. – C Weid Dec 6 '12 at 6:45
I've found another method.The idea is to find the differential equation which $f(x)$ and $f'(x)$ satisfy.It's a first-order linear ODE,solve the Cauchy problem then we get $f(x)$.Still it costs quite some pen and paper. – C Weid Dec 6 '12 at 10:42
There is a little mistake in your solution.When $t>0$ the answer is correct,the $t<0$ case need to be more careful about the residue at $u=\sqrt{-\frac{1}{t^2}-1}$. – C Weid Dec 6 '12 at 14:04
I've corrected it. – C Weid Dec 6 '12 at 14:21

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