How I can calculate following integrals for large values $\alpha$ in Mathematica:
$$ I_1 =\int_{0}^{y} \exp\left(\, -\alpha \sqrt{x(1-x)}\,\right)\, {\rm d}x $$ $$ I_2 =\int_{0}^{y} \exp\left(\, -\alpha \sqrt{x}\,\right)\, {\rm d}x $$
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1 Answer
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The second case integrates analytically, so you can do a series expansion:
series = Normal@Series[
Integrate[Exp[-alpha Sqrt[x]], {x, 0, y}], {alpha, Infinity, 3}]
E^(-alpha Sqrt[y]) (-(2/alpha^2) + (2 E^(alpha Sqrt[y]))/alpha^2 - (2 Sqrt[y])/ alpha)
Plot[{NIntegrate[Exp[-alpha Sqrt[x]], {x, 0, 1/2}] , series /. y -> 1/2 }, {alpha, 0 , 10}]
(this plot is the 2 term series )
As a purely empirical observation, your first integral appears to have the same limit form:
Plot[{
NIntegrate[Exp[-alpha Sqrt[x (1 - x)]], {x, 0, 3/4}] ,
series /. y -> y (1 - y) /. y -> 3/4 }, {alpha, 10, 50}]
answered Aug 4, 2014 at 21:21
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$\begingroup$ Thanks, I think the first one should be integrateable if we expand the integrand around $\alpha -> inf$ $\endgroup$– RoyAug 4, 2014 at 21:42
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$\begingroup$ In the wolfram alpha there is something like (Series expansion of the integral at a=∞) which I am not sure how can be called with Mathematica. $\endgroup$– RoyAug 4, 2014 at 22:10
Limit[ Integrate[ Exp[ -alpha Sqrt[x] ], {x, 0, y}] , alpha -> Infinity, Assumptions -> {y > 0}]. The first you probably need to do numerically $\endgroup$