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

Are the following integrals equal for large $\alpha$:

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

According to the answers I got from this forum they both equal to:

$$I \sim_{\alpha \sim \infty} \frac{1}{\alpha^2}- {\frac { \left( \alpha\,y+1 \right) {{\rm e}^{ -\alpha\,y}}}{{\alpha}^{2}}}.$$

However, it doesn't make sense since $\sqrt{x}$ and $\sqrt{x(1-x)}$ are very different and the plot of these functions show the difference. If they be equal then following should be equal which is not clearly equal:

$$\int_{y_1}^{y_1+dy} \exp\left(\, -\alpha \sqrt{x(1-x)}\,\right)\, {\rm d}x = \int_{y_1}^{y_1+dy} \exp\left(\, -\alpha \sqrt{x}\,\right)\, {\rm d}x$$

• Well, $f(\alpha)\sim_{\alpha\to\infty}g(\alpha)$ doesn't imply $f(\alpha)=g(\alpha)$, so what is it you're unsure about? – CuriousGuest Aug 4 '14 at 20:45
• Near $0$, the functions $\sqrt{x}$ and $\sqrt{x(1-x)}$ are quite similar. And for the asymptotics as $\alpha\to \infty$, only the values for $x$ very close to $0$ matter, provided that $y < 1$. Note that asymptotic equality is not equality. Both integrals have the same dominant terms, but they differ in the power order terms. – Daniel Fischer Aug 4 '14 at 20:46
• @CuriousGuest I mean if we accept that approximation then $I_1(y1)-I_1(y2) = I_2(y1)-I_2(y2)$ which for small y2-y1 clearly is not the case. – Hesam Aug 4 '14 at 20:54
• @DanielFischer Right, but if I evaluate I1(0,y1)−I1(0,y2) = I1(y1,y2)? Then it means the equality is more general. – Hesam Aug 4 '14 at 20:55
• To let $y_2-y_1\to 0$ you first need to fix $\alpha$, but in this case asymptotics for $\alpha\to\infty$ is irrelevant. – CuriousGuest Aug 4 '14 at 20:58