astro_enthu
Jul25-11, 08:45 PM
I want to solve following double integration. I am stuck from long time. Any intermediate solution will also be appreciated.
\int\int dvb dvl (feven+fodd)
where,
feven = (r*(vl^{2}+vb^{2}))^(-2*\beta)*(potential - 0.5*(vl^{2}+vb^{2}+vr^{2}))^(-3*\beta+5.5), where \beta,potential and vr are some constant numbers
fodd = (1-\eta)*tanh(r*vl*cos(b)/1000)*feven
Where, \eta, r,b are constant numbers. Which means I want to marginalise (feven+fodd) over vl and vb.
Integration limits for vl and vb comes from energy equation. Limits are -\sqrt{2*potential - vb^{2} - vr^{2}}<=vl<=\sqrt{2*potential - vb^{2} - vr^{2}} and -\sqrt{2*potential - vr^{2}}<=vb<=\sqrt{2*potential - vr^{2}}
Also -3.5< =\beta <1
and 0<= \eta <=2
Any help will be highly appreciated
\int\int dvb dvl (feven+fodd)
where,
feven = (r*(vl^{2}+vb^{2}))^(-2*\beta)*(potential - 0.5*(vl^{2}+vb^{2}+vr^{2}))^(-3*\beta+5.5), where \beta,potential and vr are some constant numbers
fodd = (1-\eta)*tanh(r*vl*cos(b)/1000)*feven
Where, \eta, r,b are constant numbers. Which means I want to marginalise (feven+fodd) over vl and vb.
Integration limits for vl and vb comes from energy equation. Limits are -\sqrt{2*potential - vb^{2} - vr^{2}}<=vl<=\sqrt{2*potential - vb^{2} - vr^{2}} and -\sqrt{2*potential - vr^{2}}<=vb<=\sqrt{2*potential - vr^{2}}
Also -3.5< =\beta <1
and 0<= \eta <=2
Any help will be highly appreciated