Menu Close

integral-of-e-2ax-2-




Question Number 132674 by aurpeyz last updated on 15/Feb/21
integral of e^(−2ax^2 )
$${integral}\:{of}\:{e}^{−\mathrm{2}{ax}^{\mathrm{2}} } \\ $$
Answered by Olaf last updated on 15/Feb/21
Let Ω_a (x) = ∫_0 ^x e^(−2at^2 ) dt  1) a = 0  Ω_0 (x) = ∫_0 ^x dt = x  2) a > 0  Let u = (√(2a))t  Ω_a (x) = ∫_0 ^((√(2a))x) e^(−u^2 ) .(du/( (√(2a))))  Ω_a (x) = ((√π)/( 2(√(2a))))((2/( (√π)))∫_0 ^((√(2a))x) e^(−u^2 ) du)  Ω_a (x) = (√(π/(8a))).erf((√(2a))x)  3) a < 0  Let u = i(√(−2a))t  Ω_a (x) = ∫_0 ^(i(√(−2a))x) e^(−u^2 ) .(du/( i(√(−2a))))  Ω_a (x) = ((√π)/( 2i(√(−2a))))((2/( (√π)))∫_0 ^(−i(√(2a))x) e^(−u^2 ) du)  Ω_a (x) = (√(−(π/(8a)))).((erf(i(√(−2a))x))/i)  Ω_a (x) = (√(−(π/(8a)))).erfi((√(−2a))x)
$$\mathrm{Let}\:\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} {e}^{−\mathrm{2}{at}^{\mathrm{2}} } {dt} \\ $$$$\left.\mathrm{1}\right)\:{a}\:=\:\mathrm{0} \\ $$$$\Omega_{\mathrm{0}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} {dt}\:=\:{x} \\ $$$$\left.\mathrm{2}\right)\:{a}\:>\:\mathrm{0} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{\mathrm{2}{a}}{t} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } .\frac{{du}}{\:\sqrt{\mathrm{2}{a}}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{\mathrm{2}{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{\frac{\pi}{\mathrm{8}{a}}}.\mathrm{erf}\left(\sqrt{\mathrm{2}{a}}{x}\right) \\ $$$$\left.\mathrm{3}\right)\:{a}\:<\:\mathrm{0} \\ $$$$\mathrm{Let}\:{u}\:=\:{i}\sqrt{−\mathrm{2}{a}}{t} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{i}\sqrt{−\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } .\frac{{du}}{\:{i}\sqrt{−\mathrm{2}{a}}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\frac{\sqrt{\pi}}{\:\mathrm{2}{i}\sqrt{−\mathrm{2}{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{−{i}\sqrt{\mathrm{2}{a}}{x}} {e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{−\frac{\pi}{\mathrm{8}{a}}}.\frac{\mathrm{erf}\left({i}\sqrt{−\mathrm{2}{a}}{x}\right)}{{i}} \\ $$$$\Omega_{{a}} \left({x}\right)\:=\:\sqrt{−\frac{\pi}{\mathrm{8}{a}}}.\mathrm{erfi}\left(\sqrt{−\mathrm{2a}}{x}\right) \\ $$
Commented by aurpeyz last updated on 16/Feb/21
arrrrggggg
$${arrrrggggg} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *