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Evaluate $$I=\int_{1/2014}^{2014} \dfrac{\tan^{-1}x}{x}\mathrm dx$$

$$$$I tried to solve this integral using Differentiation Under the Integral Sign. I thus redefined the integral as $$I(a)=\int_{1/2014}^{2014} \frac{\tan^{-1}(ax)}{x}\mathrm dx$$$$\Rightarrow I'(a)= \int_{1/2014}^{2014} \frac{1}{1+a^2x^2}\mathrm dx=\left(\dfrac{\tan^{-1}(ax)}{a}\right)_{1/2014}^{2014}$$

I know that there are other methods of solving this integral. However for the sake of practice, I am specifically interested in a solution involving Differentiation Under the Integral Sign.

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  • $\begingroup$ What makes you think that there will be a nice solution involving differentiation under the integral sign? $\endgroup$ – Alex Zorn May 8 '18 at 19:32
  • $\begingroup$ Try the substitution $x\mapsto\frac1x$. I see that Donald Splutterwit's answer does this. $\endgroup$ – robjohn May 8 '18 at 19:38
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    $\begingroup$ @Ishan: Differentiation under the integral sign seems to lead back to the same integral. Sometimes a particular tool doesn't work well for a particular job. That's why we have a toolbox. $\endgroup$ – robjohn May 8 '18 at 19:43
  • $\begingroup$ @Ishan The nice thing about your problem and your method is that it doesn't work. As Rob indicated Differentiation under the integral sign is a great tool not for this problem. But that does not make your post redundant. Quite the contrary! +1) $\endgroup$ – imranfat May 8 '18 at 21:49
  • $\begingroup$ Ramanujan wrote on a paper on the integral $f(x)= \int_{0}^{x}\frac{\arctan t} {t} \, dt$ in Journal of Indian Mathematical Society 7th issue, 1915, and the first thing he mentions is that it is odd function and next that $f(x) - f(1/x)=\frac{\pi}{2}\log x$. So put $x=2014$ in this relation. $\endgroup$ – Paramanand Singh Jun 8 at 11:41
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Note that \begin{eqnarray*} \tan^{-1}(x)+\tan^{-1}(1/x)=\frac{\pi}{2}. \end{eqnarray*} and \begin{eqnarray*} I= \int_{1/2014}^{2014} \frac{\tan^{-1}(x)}{x} dx =\frac{\pi}{2} \int_{1/2014}^{2014} \frac{dx}{x} + \int_{1/2014}^{2014} \frac{\tan^{-1}(1/x)}{1/x} \frac{-dx}{x^2}. \end{eqnarray*} Now substitute $u=1/x$ in the latter integral to obtain $-I$. ... should be a doddle from here ?

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    $\begingroup$ @Ishan: Sometimes an integral is much easier to do in another way. $\endgroup$ – robjohn May 8 '18 at 19:41
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This is too long for a comment; however, this might shed some light on why differentiation under the integral sign leads back to the same integral: $$ \begin{align} &\int_0^1\left(\color{#C00}{\frac{\mathrm{d}}{\mathrm{d}a}\int_{1/2014}^{2014}\frac{\tan^{-1}(ax)}x\,\mathrm{d}x}\right)\mathrm{d}a\tag1\\ &=\int_0^1\left(\frac{\mathrm{d}}{\mathrm{d}a}\int_{a/2014}^{2014a}\frac{\tan^{-1}(x)}x\,\mathrm{d}x\right)\mathrm{d}a\tag2\\ &=\int_0^1\left(\color{#C00}{\frac{\tan^{-1}(2014a)}a-\frac{\tan^{-1}\left(\frac{a}{2014}\right)}a}\right)\mathrm{d}a\tag3\\ &=\int_0^{2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a-\int_0^{1/2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a\tag4\\ &=\int_{1/2014}^{2014}\frac{\tan^{-1}(a)}a\,\mathrm{d}a\tag5 \end{align} $$ Explanation:
$(1)$: the integral for $a=0$ is $0$
$(2)$: substitute $x\mapsto x/a$
$(3)$: Fundamental Theorem of Calculus
$(4)$: substitute $a\mapsto a/2014$ on the left and $a\mapsto2014a$ on the right
$(5)$: subtract the integrals

No matter how you slice it, the red expressions are the same, and you will run back into the original integral.

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Thus $$I' (a) =\int_{2014^{-1}}^{2014} \frac{1}{1+a^2 x^2 } dx =a^{-1}\int_{2014^{-1}a}^{2014 a}\frac{1}{1+t^2 } dt=a^{-1}\arctan t|_{2014^{-1}a}^{2014 a}$$

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    $\begingroup$ This seems to lead back to the same integral. $\endgroup$ – robjohn May 8 '18 at 19:41

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