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What is the value of following integral? $$\int_{1/2014}^{2014}\frac{\tan^{-1}x} x \, dx$$

I am having problem evaluating this.

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    $\begingroup$ the result should be $$\frac{1}{2} \pi \log (2014)$$ $\endgroup$ – Dr. Sonnhard Graubner May 13 '17 at 16:23
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    $\begingroup$ By integration by parts: $$\int_\frac{1}{\text{n}}^\text{n}\frac{\arctan\left(x\right)}{x}\space\text{d}x=$$ $$\arctan\left(\text{n}\right)\ln\left|\text{n}\right|-\arctan\left(\frac{1}{\text{n}}\right)\ln\left|\frac{1}{\text{n}}\right|-\int_\frac{1}{\text{n}}^\text{n}\frac{\ln\left|x\right|}{1+x^2}\space\text{d}x$$ $\endgroup$ – Jan May 13 '17 at 16:29
  • $\begingroup$ It may be useful to remember that when $x>0$ then $\arctan x + \arctan \dfrac 1 x = \dfrac \pi 2. \qquad$ $\endgroup$ – Michael Hardy May 13 '17 at 16:50
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    $\begingroup$ I'm voting to close this question as off-topic because it is nothing more than a problem statement, and the OP wants the answer. There is no signs of effort from the asker, no context provided, and this user has exhibited a long streak of asking problem statements without context, showing no effort. $\endgroup$ – amWhy May 14 '17 at 19:37
  • $\begingroup$ @amWhy Very glad to hear that you're judging me, well close it I don't care, there are many good people here, certainly you're not one of them, maybe you're more experienced but that doesn't give you right to insult others, trying to make me seem troll here, while you're the one. If I was able to solve all of this I wouldn't have asked it on here and if I like showing no efforts as you say I wouldn't solve these question at first place, why give a headache even seeing them right? Go enjoy your virtual ego and fame. $\endgroup$ – Iti Shree May 14 '17 at 20:00
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Let $x = \frac{1}{u}, dx = -\frac{1}{u^2} du$. Then $$I=\int_{1/2014}^{2014} \frac{\tan^{-1}(x)}{x}dx = \int_{2014}^{1/2014} -\frac{\tan^{-1}(1/u)}{u}du$$ For positive $u$, we have $\tan^{-1}(1/u) = \frac{\pi}{2}-\tan^{-1}(u)$, and so $$= \int_{1/2014}^{2014} \frac{\pi}{2u} - \frac{\tan^{-1}(u)}{u}\, du = \int_{1/2014}^{2014} \frac{\pi}{2u}\, du -I$$ and so $$2I = \frac{\pi}{2}\int_{1/2014}^{2014} \frac{1}{u}\, du = \pi \ln(2014)$$ giving us $$I = \frac{\pi}{2}\ln(2014)$$

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I suggest to try the substitution $y=1/x$ and use the fact that $\tan^{-1}x+\tan^{-1}(1/x)=\dfrac{\pi}{2}$ for $x>0$.

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  • $\begingroup$ I did that but then I forgot that $tan^{-1}\frac{1}{u}$ could be written as $\frac{\pi}{2}-tan^{-1}(u)$ $\endgroup$ – Iti Shree May 13 '17 at 16:41
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\begin{align} w & = \frac 1 x \\[10pt] dw & = \frac{-dx}{x^2}, \text{ so } \frac{dw} w = \frac{-dx} x. \\[20pt] \int_{1/2014}^{2014} \frac{\arctan x} x \, dx & = \int_{2014}^{1/2014} \frac{\arctan(1/w)}{w} (-dw) \\[10pt] & = \int_{2014}^{1/2014} \frac{\frac \pi 2 - \arctan w} w \, (-dw) \\[10pt] & = \int_{1/2014}^{2014} \frac \pi {2w} \, dw - \int_{1/2014}^{2014} \frac {\arctan w} w \, dw. \\[10pt] \text{So } I & = \int_{1/2014}^{2014} \frac \pi {2w} \, dw - I, \\[10pt] \text{and thus } 2I & = \int_{1/2014}^{2014} \frac \pi {2w}\, dw, \end{align} and then divide both sides by $2$.

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