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I want to integrate $$\int_0^{L/2}\int_0^{L/2} \frac{1}{\sqrt{x^2+y^2+z^2}} \,dx\,dy\ $$ It looks simple, but I can't integrate. I calculate inner integral, so the result is $$\int_0^{L/2} \left[ \ln(L/2+\sqrt{(L/2)^2+y^2+z^2}) - \ln(\sqrt{y^2+z^2}) \right]\,dy $$ but I can't finish this integral. What should I do?

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    $\begingroup$ What are the limits of the integral in the third variable? Is there one $\int_0^{L/2}$ missing? $\endgroup$ – mickep Aug 22 '17 at 6:34
  • $\begingroup$ @mickep I'm sorry. I rewrite the integral. $\endgroup$ – Ted Aug 22 '17 at 7:09
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I will just give some outlines. First, write your integral as a double integral, use symmetry to write it for half the square only, and then change the coordinates to polar coordinates. You will end up with $$ I=2\iint_D\frac{r}{\sqrt{r^2+z^2}}\,dr\,d\phi $$ where $$ D=\{(r,\phi)~|~0<r<L/(2\cos\phi),\ 0<\phi<\pi/4\}. $$ Integrating in $r$ becomes easy, and you find that $$ I=2\int_0^{\pi/4}\sqrt{z^2+\frac{L^2}{4\cos^2\phi}}-|z|\,d\phi. $$ Let $u=\tan\phi$, do some calculations, and you will find that a primitive to the integrand is given by $$ z\arctan\biggl(\frac{z\tan\phi}{\sqrt{z^2+L^2/(4\cos^2\phi)}}\biggr)+\frac{L}{2}\mathop{\text{artanh}}\biggl(\frac{(L/2)\tan\phi}{\sqrt{z^2+L^2/(4\cos^2\phi)}}\biggr)-|z|\phi. $$ Inserting limits, you end up with

$$I=2z\arctan\biggl(\frac{z}{\sqrt{z^2+L^2/2}}\biggr)+L\mathop{\text{artanh}}\biggl(\frac{L/2}{\sqrt{z^2+L^2/2}}\biggr)-\frac{\pi}{2}|z|.$$

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  • $\begingroup$ Although I'm in trouble with some calculations, Thank you! $\endgroup$ – Ted Aug 23 '17 at 5:42
  • $\begingroup$ I'm sorry, but I can't calculate after u=tan(phi). Could you give me a hint? $\endgroup$ – Ted Aug 24 '17 at 13:18
  • $\begingroup$ Next substitution I made was $$v=\frac{u}{\sqrt{z^2+L^2(1+u^2)/4}}$$ $\endgroup$ – mickep Aug 24 '17 at 13:29

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