# Calculate integral in limit of very large coefficient

How I can calculate following integrals for large values $\alpha$ in Mathematica:

$$I_1 =\int_{0}^{y} \exp\left(\, -\alpha \sqrt{x(1-x)}\,\right)\, {\rm d}x$$ $$I_2 =\int_{0}^{y} \exp\left(\, -\alpha \sqrt{x}\,\right)\, {\rm d}x$$

• The second is zero: Limit[ Integrate[ Exp[ -alpha Sqrt[x] ], {x, 0, y}] , alpha -> Infinity, Assumptions -> {y > 0}]. The first you probably need to do numerically Aug 4, 2014 at 21:13
• Thanks, Actually I dont want the limit but expansion considering $\alpha$ is very large. (i.e $2/ \alpha^2$). Do yo know how?
– Roy
Aug 4, 2014 at 21:16

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}] • Thanks, I think the first one should be integrateable if we expand the integrand around $\alpha -> inf$
– Roy
Aug 4, 2014 at 21:42
• 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.
– Roy
Aug 4, 2014 at 22:10