Question Number 98826 by bramlex last updated on 16/Jun/20
$${Given}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dx}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{2}{a}} \\ $$$${find}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:? \\ $$
Commented by bemath last updated on 16/Jun/20
![(d/da) [∫_0 ^∞ (dx/(a^2 +x^2 )) ] = (d/da) [(π/(2a)) ] −2a ∫_0 ^∞ (dx/((a^2 +x^2 )^2 )) = ((−π)/(2a^2 )) ∫_0 ^∞ (dx/((a^2 +x^2 )^2 )) = (π/(4a^3 )) (d/da) [ ∫_0 ^∞ (dx/((a^2 +x^2 )^2 )) ] = (d/da) [(π/(4a^3 )) ] −4a ∫_0 ^∞ (dx/((a^2 +x^2 )^3 )) = ((−3π)/(4a^4 )) ∴ ∫_0 ^∞ (dx/((a^2 +x^2 )^3 )) = ((3π)/(16a^5 )) ■](Q98829.png)
$$\frac{\mathrm{d}}{\mathrm{da}}\:\left[\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{dx}}{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\:\right]\:=\:\frac{\mathrm{d}}{\mathrm{da}}\:\left[\frac{\pi}{\mathrm{2a}}\:\right] \\ $$$$−\mathrm{2}{a}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{−\pi}{\mathrm{2a}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{4a}^{\mathrm{3}} } \\ $$$$\frac{\mathrm{d}}{\mathrm{da}}\:\left[\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\right]\:=\:\frac{\mathrm{d}}{\mathrm{da}}\:\left[\frac{\pi}{\mathrm{4a}^{\mathrm{3}} }\:\right] \\ $$$$−\mathrm{4a}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\:\frac{−\mathrm{3}\pi}{\mathrm{4a}^{\mathrm{4}} } \\ $$$$\therefore\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{dx}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\:\frac{\mathrm{3}\pi}{\mathrm{16a}^{\mathrm{5}} }\:\blacksquare \\ $$
Commented by bramlex last updated on 16/Jun/20
$${greattt} \\ $$