Saddle point approximation gives a null result

Question

So I want to compute the following integral $$I=\int_0^1 x\sqrt{1-x}\exp \left(a^2x^2\right) dx$$ where $a>>1$. If we try to do a Saddle point approximation \begin{align} I&=\int_0^1 f(x)\exp \left(a^2g(x)\right)\\ &\approx f(x_0)\exp\left(a^2g(x_0)\right)\sqrt{\frac{2\pi}{-a^2g''(x_0)}}\left(1+o\left(1/a^2\right)\right) \end{align} where $g(x_0)=0$. In the case of $I$, we have $g(x)=x^2$ and $f(x)=x\sqrt{1-x}$ and $x_0=0$. The problem is that $f(x_0)=0$ so this gives us $I=0$. How can one get around such a problem?

Answer 1

The problem you have is that the prefactor ($f$) vanishes at the saddle point. One possibility here is to integrate by parts. Note that $$ x\exp(a^2x^2)dx=\frac{1}{2a^2}d\left(\exp(a^2x^2)\right); $$ so $$ \begin{eqnarray} \int_{0}^{1}x\sqrt{1-x}\exp(a^2x^2)dx&=&\frac{1}{2a^2}\int_{0}^{1}\sqrt{1-x}\cdot d\left(\exp(a^2x^2)\right) \\ &=&\frac{1}{2a^2}\sqrt{1-x}\exp(a^2x^2)\big\vert_{0}^{1} + \frac{1}{4a^2}\int_{0}^{1}\frac{\exp(a^2 x^2)}{\sqrt{1-x}}dx \\ &=&-\frac{1}{2a^2}+\frac{1}{4a^2}\int_{0}^{1}\frac{\exp(a^2x^2)}{\sqrt{1-x}}dx. \end{eqnarray} $$ This is exact (up to algebra mistakes I may have made :)), and you can apply the saddle point approximation to the remaining integral.

It may seem like a lucky accident that integration by parts can be applied. However, quite generally you can expand $f(x)$ and $g(x)$ as power series around the saddle point $x=x_0$, and when the constant term $f(x_0)$ vanishes, you can integrate by parts. Sometimes this needs to be done repeatedly: for instance, if $f(x)\sim (x-x_0)^2$, you need to integrate by parts twice.

Answer 2

Well, let's first review:

• $$\sqrt{1-x}=\sum_{\text{n}\ge0}\left(-1\right)^\text{n}x^\text{n}\binom{\frac{1}{2}}{\text{n}}\tag1$$
• $$\exp(x)=\sum_{\text{k}\ge0}\frac{x^\text{k}}{\text{k}!}\tag2$$

Let's define the following integral:

$$\mathcal{I}_\alpha:=\int_0^1x\sqrt{1-x}\exp\left(\alpha^2x^2\right)\space\text{d}x\tag3$$

Using $(1)$ and $(2)$, we can write:

$$\mathcal{I}_\alpha=\int_0^1x\cdot\left(\sum_{\text{n}\ge0}\left(-1\right)^\text{n}x^\text{n}\binom{\frac{1}{2}}{\text{n}}\right)\cdot\left(\sum_{\text{k}\ge0}\frac{\left(\alpha^2x^2\right)^\text{k}}{\text{k}!}\right)\space\text{d}x=$$ $$\sum_{\text{n}\ge0}\left(-1\right)^\text{n}\binom{\frac{1}{2}}{\text{n}}\sum_{\text{k}\ge0}\frac{\alpha^{2\text{k}}}{\text{k}!}\int_0^1x\cdot\left(x^\text{n}\right)\cdot\left(x^{2\text{k}}\right)\space\text{d}x=$$ $$\sum_{\text{k}\ge0}\sum_{\text{n}\ge0}\left(-1\right)^\text{n}\binom{\frac{1}{2}}{\text{n}}\frac{\alpha^{2\text{k}}}{\text{k}!}\underbrace{\int_0^1x^{1+\text{n}+2\text{k}}\space\text{d}x}_{\text{I}}\tag4$$

And it is not hard to see:

$$\text{I}=\left[\frac{x^{2+\text{n}+2\text{k}}}{2+\text{n}+2\text{k}}\right]_0^1=\frac{1^{2+\text{n}+2\text{k}}}{2+\text{n}+2\text{k}}-\frac{0^{2+\text{n}+2\text{k}}}{2+\text{n}+2\text{k}}=\frac{1}{2+\text{n}+2\text{k}}\tag5$$

So, we end up with:

$$\mathcal{I}_\alpha=\sum_{\text{k}\ge0}\sum_{\text{n}\ge0}\left(-1\right)^\text{n}\binom{\frac{1}{2}}{\text{n}}\frac{\alpha^{2\text{k}}}{\text{k}!}\frac{1}{2+\text{n}+2\text{k}}\tag6$$