Prove an integral to be $o(h)$ [closed]

I've got an expression (if not wrong, see my last post for original question) $$\int_{y=-x}^x e^{\mu y-\frac{1}{2}\mu^2 h}\frac{1}{\sqrt{2\pi h}} (e^{-\frac{(y-2x)^2}{2h}} +e^{-\frac{(y+2x)^2}{2h}}-e^{-\frac{(y-4x)^2}{2h}}-e^{-\frac{(y+4x)^2}{2h}})\,dy$$ and now am asked to prove it's of $o(h)$, when $h\to 0$. But I have no idea to proceed.

Any suggestion will be appreciated.

Solution 1:

Notice that your integral only depends upon $y$ so we can take out the constant parts, id est:

$$\frac{e^{-\frac{1}{2} h\mu^2}}{\sqrt{2\pi h}} \int_{-x}^x e^{\mu y} \left(\text{all the exponentials}\right)\ \text{d}y$$

Your integral is now a rather standard Gaussian integral and you can split into the four pieces.. For example, let's evaluate the first piece which is

$$\frac{e^{-\frac{1}{2} h\mu^2}}{\sqrt{2\pi h}} \int_{-x}^x e^{\mu y} e^{-\frac{(y-2x)^2}{2h}}\ \text{d}y$$

Expand the square, and use the property of exponential to get

$$\frac{e^{-\frac{1}{2} h\mu^2}}{\sqrt{2\pi h}} \int_{-x}^x e^{-\frac{y^2}{2h} + y\left(\mu + \frac{2x}{h}\right) - \frac{2x^2}{h}}\ \text{d}y$$

Now this is special Integral which is of the form

$$\int e^{-ay^2 + by + c}\ \text{d}y$$

Whose solution can be expressed in terms of a Special Function called the Error Function:

$$\int e^{-ay^2 + by + c}\ \text{d}y = \frac{\sqrt{\pi } e^{\frac{b^2}{4 a}+c} \text{erf}\left(\frac{2 a y-b}{2 \sqrt{a}}\right)}{2 \sqrt{a}}$$

Just to give you some help: using

$$a = \frac{1}{2h} ~~~~~~~ b = \mu + \frac{2x}{h} ~~~~~~~ c = \frac{2x^2}{h}$$

(where $a$ was already negative in the integral so I did not put the minus sign now)

you obtain for the first piece:

$$\frac{e^{-\frac{1}{2} h\mu^2}}{\sqrt{2\pi h}} \sqrt{\frac{\pi }{2}} \sqrt{h} e^{\frac{4 x^2}{h}+\frac{h \mu ^2}{2}+2 x \mu } \left(\text{erf}\left(\frac{h \mu +3 x}{\sqrt{2} \sqrt{h}}\right)-\text{erf}\left(\frac{h \mu +x}{\sqrt{2} \sqrt{h}}\right)\right)$$

That is,

$$\frac{1}{2} e^{2 x \left(\frac{2 x}{h}+\mu \right)} \left(\text{erf}\left(\frac{h \mu +3 x}{\sqrt{2} \sqrt{h}}\right)-\text{erf}\left(\frac{h \mu +x}{\sqrt{2} \sqrt{h}}\right)\right)$$

Now it's about hints and ideas

The formal asymptotic for the Error Function is

$$\text{erf}(z) \sim \frac{e^{-z^2}}{z\sqrt{\pi}} \sum_{k = 0}^{+\infty} (-1)^k \frac{(2k-1)!!}{(2z^2)^k}$$

Where in your case $z$ is the whole argument. With this in mind, after few calculations (where I stopped the series after the second term and I neglected the $\mu$ term for it is not asymptotically relevant), we get

$$\frac{1}{2} e^{2 x \left(\frac{2 x}{h}+\mu \right)} \left(\text{erf}\left(\frac{h \mu +3 x}{\sqrt{2} \sqrt{h}}\right)-\text{erf}\left(\frac{h \mu +x}{\sqrt{2} \sqrt{h}}\right)\right) \sim \frac{1}{x}\sqrt{\frac{h}{2\pi}}\left[\frac{1}{3}e^{7x^2/4h} - e^{15x^2/4h}\right]$$

Now this doesn't look like an $o(h)$ for $h\to 0$. But this is just the first one term in your integral. There are three others to be calculated. I gave you some hints and tools, you might try to have fun and proceed.