Hint given: Write the integrand as an integral.
I'm supposed to do this as double integration.
My attempt:
$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$
$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$
$$= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}$$
$$= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
$$= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
Carrying out this integration, I got, $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]$$
But I'm supposed to get $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5$$
Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.
$\endgroup$ 33 Answers
$\begingroup$Use the change of variables $y = xt.$
$$I = \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \, \mathrm{d}x} {y^2+1}\\=\int_0^2 \int_1^{\pi } \frac { x} {x^2t^2+1}\mathrm{d}t \, \mathrm{d}x\\=\int_1^{\pi} \int_0^{2 } \frac { x} {x^2t^2+1}\mathrm{d}x \, \mathrm{d}t\\=\int_1^{\pi} \frac{\ln(1+4t^2)}{2t^2} \mathrm{d}t$$
Now use integration by parts.
$$I = -\left.\frac{\ln(1+4t^2)}{2t}\right|_1^{\pi}+4\int_1^{\pi}\frac1{1+4t^2} \, dt$$
$\endgroup$ $\begingroup$A more straight forward approach uses integration by parts.
Define: \begin{align} & I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2} \end{align}
using partial fraction this reads: \begin{align} I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc} \end{align}
taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done
$\endgroup$ $\begingroup$For the sake of an alternative approach, recall the formula for the integral of an inverse function$$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-F\circ f^{-1}(x)$$Where $F'(x)=f(x)$. Plugging in $f(x)=\tan(x)$,$$I=\int\arctan(x)\mathrm{d}x=x\arctan(x)-\int_0^{\arctan(x)}\tan(t)\mathrm{d}t$$Then recall that $$(-\log|\cos(x)|)'=\tan(x)$$So we have$$I=x\arctan(x)+\log|\cos(\arctan(x))|$$then using trig,$$I=x\arctan(x)-\frac12\log(x^2+1)$$So $$I_1=\int_0^2\arctan(x)\mathrm{d}x=2\arctan2-\frac12\log5$$And $$ \begin{align} I_2=&\int_0^2\arctan(\pi x)\mathrm{d}x\\ =&\frac1\pi\int_0^{2\pi}\arctan(x)\mathrm{d}x\\ =&\frac1\pi\bigg(2\pi\arctan2\pi-\frac12\log(4\pi^2+1)\bigg)\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)\\ \end{align} $$So $$ \begin{align} \int_0^2(\arctan\pi x-\arctan x)\mathrm{d}x=&I_2-I_1\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)-2\arctan2+\frac12\log5\\ =&2\arctan2\pi-\frac1{2}\log\sqrt[\pi]{4\pi^2+1}-2\arctan2+\frac12\log5\\ =&2(\arctan2\pi-\arctan2)+\frac12\log\frac{5}{\sqrt[\pi]{4\pi^2+1}}\\ \end{align} $$
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