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Question-83559

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Question Number 83559 by Jidda28 last updated on 03/Mar/20
Commented by Jidda28 last updated on 03/Mar/20
help me out pls.
helpmeoutpls.
Answered by mind is power last updated on 04/Mar/20
=∫_0 ^(+∞) e^(−ax^2 ) .x^(2m−1) dx.∫_0 ^∞ e^(−by^2 ) y^(2n−1) dy u=x^2 ,⇒du=2xdx,v=y^2 ⇒ =∫_0 ^(+∞) (u^(m−1) /2)e^(−au) du.∫_0 ^∞ e^(−bv) .(v^(n−1) /2)dv =(1/2)∫_0 ^(+∞) (t^(m−1) /a^(m−1) ).e^(−t) .(dt/a).(1/2)∫_0 ^(+∞) ((e^(−t) .t^(n−1) )/b^(m−1) ).(dt/b) =(1/(4(ab)^m ))∫_0 ^(+∞) t^(m−1) e^(−t) dt.∫_0 ^(+∞) t^(n−1) e^(−t) dt =((Γ(m).Γ(n))/(4(ab)^m ))
=∫0+∞e−ax2.x2m−1dx.∫0∞e−by2y2n−1dyu=x2,⇒du=2xdx,v=y2⇒=∫0+∞um−12e−audu.∫0∞e−bv.vn−12dv=12∫0+∞tm−1am−1.e−t.dta.12∫0+∞e−t.tn−1bm−1.dtb=14(ab)m∫0+∞tm−1e−tdt.∫0+∞tn−1e−tdt=Γ(m).Γ(n)4(ab)m
Commented by Jidda28 last updated on 04/Mar/20
thank you sir
thankyousir

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