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How to evaluate the following integral $$\int_0^\infty\frac{dl}{(r^2+l^2)^{\large\frac32}}$$

The solution is supposed to look like this, unfortunately I can't derive it. $$ \left[\frac{l}{r^2\sqrt{r^2+l^2}}\right]_{l=0}^\infty $$

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  • $\begingroup$ Wolphram Alpha premium account -> input integral -> step by step solution. This is not a physics problem. $\endgroup$ – M.Herzkamp Jul 28 '14 at 11:39
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    $\begingroup$ Please use $\LaTeX$ for all formulas. The "formula" in the image is just unreadable. $\endgroup$ – Ruslan Jul 28 '14 at 11:57
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    $\begingroup$ Did you try substitution? E.g. $y = l^2+r^2$? $\endgroup$ – flawr Jul 28 '14 at 12:02
  • $\begingroup$ Since $ l$ is the variable of integration, substitute $l = r\tan \theta$. $\endgroup$ – Namaste Jul 28 '14 at 12:04
  • $\begingroup$ If you know the solution to the indeterminate integral, then differentiate and see what happens. $\endgroup$ – Arthur Jul 28 '14 at 12:14
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Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes $$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$

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All integrals of the form $\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx$ can be expressed in terms of trigonometric functions

as follows: first, let $x=at$, then $u=\dfrac1{t^n+1}$. Now recognize the expression of the beta function in

the new integral, and finally use Euler's reflection formula for the $\Gamma$ function in order to arrive at

the desired result.

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$$ \int\frac{dl}{\left(r^2+l^2\right)^{3/2}} = -\frac{1}{r}\frac{d}{dr}\int\frac{1}{\left(r^2+l^2\right)^{1/2}}dl $$ change of variables $v=l/r$ we find

$$ -\frac{1}{r}\frac{d}{dr}\frac{1}{r}\int^{\infty}_{0}\frac{1}{\left(1+v^2\right)^{1/2}}rdv = -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} $$ $$ -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+v^2}}\frac{dv}{dr}\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+\left(\frac{l}{r}\right)^2}}\left(\frac{-l}{r^2}\right)\right]^{\infty}_{0} $$ which leads to your result.

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$$I=\int_0^\infty \frac{dx}{(x^2+r^2)^{3/2}}$$ Let $xr=\tan u$. Therefore $dx=\frac1r\sec^2u\ du$: $$I=\frac1r\int_0^{\pi/2}\frac{\sec^2u\ du}{(r^2\tan^2u+r^2)^{3/2}}$$ $$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{(\tan^2u+1)^{3/2}}$$ $$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{(\sec^2u)^{3/2}}$$ $$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{\sec^3u}$$ $$I=\frac1{r^2}\int_0^{\pi/2}\frac{du}{\sec u}$$ $$I=\frac1{r^2}\int_0^{\pi/2}\cos u\ du$$ $$I=\frac1{r^2}$$

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