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$\int_{1}^{4} \int_{-1}^{2z} \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy \; dx\; dz$

Consider $ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy = $ $\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} - \frac{y}{x^2+y^2} \;dy$

$\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} \; dy = \arctan(\sqrt{3}) = \pi/3$

$x^2 + y^2 = t \Rightarrow y \; dy = dt/2$

$ \int_{0}^{\sqrt{3}x} \frac{y}{x^2 +y^2} \;dy = \int_{x^2}^{3+x^2} \frac{dt}{t} = \frac{\ln(3+x^2)}{2} - \ln(x)$

$ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy = \pi/3 + \ln(x) - \frac{\ln(3+x^2)}{2}$

Now it will be very lengthy and difficult to integrate this with respect to x and z. So I am stuck here. Is there any easy method to solve such questions ?

eranreches
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So Lo
  • 1,567

1 Answers1

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$$\int\frac{x-y}{x^2+y^2}\,dy=\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log(x^2+y^2).$$ Therefore, your integral becomes $$\begin{align}&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\ =&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4)-\frac{1}{2}\log(x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\ =&\int_1^4\int_{-1}^{2z}\left(\frac{\pi}{3}-\log(2)\right)\,dx\,dz\\ =&\left(\frac{\pi}{3}-\log(2)\right)\int_1^4(2z+1)dz\\ =&18\left(\frac{\pi}{3}-\log(2)\right)\\ =&6\pi-18\log 2 \end{align}$$

Your mistake was in calculating $\int\frac{y}{x^2+y^2}\,dy.$ Note that $$\int\frac{y}{x^2+y^2}\,dy=\frac{1}{2}\log(x^2+y^2).$$

A. Goodier
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