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This question already has an answer here:

Find $$\int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm dx$$

The hint is also given : Re-write the Integrand as an Integral

I think we have to Re-write this single integral as a double integral and then do something.

Any one any ideas

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marked as duplicate by Carl Mummert, Simon S, Hakim, Jimmy R., saz Dec 1 '14 at 19:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ When you tried the hint, what did you get? Where were your particular difficulties? $\endgroup$ – Clayton Nov 26 '14 at 18:34
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    $\begingroup$ Technically this can be done with no hint: $\int \arctan x = x \arctan x - \frac{1}{2} \log \left ( x^2+1 \right ) + C$, by a straightforward integration by parts. $\endgroup$ – Ian Nov 26 '14 at 18:45
  • $\begingroup$ math.stackexchange.com/questions/1005422/… $\endgroup$ – tired Nov 26 '14 at 18:46
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I think the hint is

$$\arctan\pi x -\arctan x=\int_x^{\pi x}\frac{du}{1+u^2}$$

Be careful with how you handle the variables.

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    $\begingroup$ $\int_{a(x)}^{b(x)} f(x)dx$, wow, that's some ugly notation $\endgroup$ – Alexandre Halm Nov 26 '14 at 18:42
  • $\begingroup$ While we all know what you mean, you really should write $\int_x^{\pi x} \frac{dy}{1+y^2}$ (or whatever other variable you like). $\endgroup$ – Ian Nov 26 '14 at 18:43
  • $\begingroup$ You are right, @ian. Thanks, edited. $\endgroup$ – Timbuc Nov 26 '14 at 19:03
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We'll generalize it a bit$$I(\alpha )=\int_0^2 \arctan(\alpha x)\, \mathrm ddx$$ And we have $I(0)=0$

$$I'(\alpha )=\int_0^2 \frac{x}{1+\alpha ^2x^2}\, \mathrm dx=\left[\frac{1}{2\alpha ^2}\ln (1+\alpha ^2x^2)\right]_0^2=\frac{\ln(1+4\alpha ^2)}{2\alpha ^2}$$

$$\begin{align} I(\alpha )&=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }+c\\ &=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }\\ \end{align}$$

$$\large I(\alpha )=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }$$

And Finally use

$$\int_0^2\arctan(\pi x)-\arctan( x)\, \mathrm dx=I(\pi)-I(1)$$

To get

$$\begin{align} \int_0^2\arctan(\pi x)-\arctan( x)\, \mathrm dx&=\frac{1}{2}\left[4\arctan(2\pi )-\frac{\ln(4\pi^2+1)}{\pi}-4\arctan(2)+\ln(5)\right]\\ &\approx 0.8274\\ \end{align}$$

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$\displaystyle \arctan\pi x -\arctan x=\int_x^{\pi x}\frac{du}{1+u^2}=\int_1^{\pi} \dfrac{x}{1+x^2u^2}du$

Therefore:

$I=\displaystyle \int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm{ dx}=\int_0^2\int_1^{\pi} \dfrac{x}{1+x^2u^2}\mathrm{ dx}\mathrm{ du}=\int_1^{\pi}\Big(\int_0^2\dfrac{x}{1+x^2u^2}\mathrm{ dx}\Big)\mathrm{ du}$

$\displaystyle \int_0^2\dfrac{x}{1+x^2u^2}\mathrm{ dx}=\dfrac{1}{2u^2}\Big[\ln(1+x^2u^2)\Big]_0^2=\dfrac{\ln(1+4u^2)}{2u^2}$

Therefore:

$I=\displaystyle \int_1^{\pi} \dfrac{\ln(1+4u^2)}{2u^2}\mathrm{ du}=\Big[-\dfrac{1}{2u}\ln(1+4u^2)\Big]_1^{\pi}+\int_1^{\pi}\dfrac{8u}{1+4u^2}\times\dfrac{1}{2u}\mathrm{ du}$

$I=\displaystyle \dfrac{1}{2}\ln 5-\dfrac{1}{2\pi}\ln(1+4\pi^2)+\Big[2\arctan (2u)\Big]_1^{\pi}$

$I=\dfrac{1}{2}\ln 5-\dfrac{1}{2\pi}\ln(1+4\pi^2)+2\arctan(2\pi)-2\arctan 2$

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