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I can't figure out how to compute the following integral: $$\int_{-\infty}^{+\infty}\frac{\mathrm{d}z}{\sqrt{(x^2+y^2+z^2)^3}}$$ I think I should do some substitution, but I didn't figure it out. Can you please give me a hint?

According to Mathematica the result should be $$\left[\frac{z}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}\right]_{-\infty}^{+\infty}=\frac{2}{x^2+y^2}$$

Thank you very much for your effort.

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  • $\begingroup$ $z = \sqrt{x^2+y^2}\sinh t$ $\endgroup$ – Daniel Fischer May 6 '14 at 19:56
  • $\begingroup$ thanks :-) I'll try that if you think this is the way to go. If I'm not wrong, this way I'll have to integrate $\frac{1}{(\sinh t)^{3/2}}$. Well I was afraid of that but I think I see how it should be done now $\endgroup$ – Tom83B May 6 '14 at 19:59
  • $\begingroup$ Another option is $z=\sqrt{x^2+y^2} \tan{t}$. $\endgroup$ – Ron Gordon May 6 '14 at 20:00
  • $\begingroup$ @Tom83B You get something simpler, fortunately. $\endgroup$ – Daniel Fischer May 6 '14 at 20:01
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Subbing $z=\sqrt{x^2+y^2} \tan{t}$, the integral becomes

$$\frac1{x^2+y^2}\int_{-\pi/2}^{\pi/2} dt \, \frac{\sec^2{t}}{\sec^3{t}} = \frac1{x^2+y^2}\int_{-\pi/2}^{\pi/2} dt \,\cos{t} = \frac{2}{x^2+y^2}$$

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  • $\begingroup$ What in the integrand hints at this substitution? $\endgroup$ – Ali Caglayan May 7 '14 at 8:02
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Let $z=\sqrt{x^2+y^2}\tan\theta$, then $dz=\sqrt{x^2+y^2}\sec^2\theta\ d\theta$. $$ \begin{align} \int_{-\infty}^{+\infty}\frac{dz}{\sqrt{(x^2+y^2+z^2)^3}}&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta\ d\theta}{\sqrt{((x^2+y^2)+(x^2+y^2)\tan^2\theta)^3}}\\ &=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta}{\sqrt{(x^2+y^2)^3}\sqrt{(1+\tan^2\theta)^3}}d\theta\\ &=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sec^2\theta\ }{\sqrt{(x^2+y^2)^2}\sec^3\theta}d\theta\\ &=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\cos\theta\ }{x^2+y^2}d\theta\\ &=\left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}\\ &=\frac{2}{x^2+y^2}. \end{align} $$ Since $\tan\theta=\dfrac{z}{\sqrt{x^2+y^2}}$, then $\sin\theta=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$. Hence $$ \left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}=\left.\frac{1}{x^2+y^2}\cdot\dfrac{z}{\sqrt{x^2+y^2+z^2}}\right|_{-\infty}^{\infty}=\frac{2}{x^2+y^2}. $$

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