How to evaluate this derivative of a double integral?

Question

I need to do a maclaurin series to first order in $\alpha$ for

$$ Q = \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0}{2}x^2 + \alpha x^4 + \frac{p^2}{2m} \right)\frac{1}{kT}} \text{d}p\text{d}x $$

We have that

$$ Q \approx Q\rvert_{a=0} + a \frac{dQ}{da}\rvert_{a=0} $$

I need to evaluate the derivative of $Q$, but this becomes pretty complicated, I think. The derivative is with respect to a different parameter than the integral variables. And I'm not able to evaluate the integrals, as I end up with a product of two exponential functions.

$$ \begin{align} Q &= \iint \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} e^{-\frac{p^2}{2mkT}} \text{d}p\text{d}x \\ &= \int \limits_{-\infty}^{\infty} e^{-\frac{p^2}{2mkT}} \text{d}p \int \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \end{align} $$

Besides, if I could do the integrals, then why the need for the maclaurin series? So how could I continue?

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Following Laray's suggestion, I end up with

$$ \frac{\text{d}Q}{\text{d}a} = - \frac{4a}{kT} \int\limits_{-\infty}^{\infty} e^{- \frac{p^2}{2mkT}} \text{d}p \int\limits_{-\infty}^{\infty}e^{- \frac{k_0x^2}{2kT}} e^{- \frac{ax^4}{kT}} \text{d}x $$

This should be evaluated at $a=0$, so the whole expression vanishes?

Answer 1

You want $\frac{dQ}{da}$ right? You are right in observing, that here actually is no double integral, but two seperate 1D-Integrations. However the Integrands are of the form $e^{-x^2}$ and therefore no closed form exists (Gaussian normal Distribution).

That does, however, not prevent you from taking the derivative, numeric evaluation comes in later.

The first integral is constant w.r.t to $a$ and therefore not affected by derivation. In the second term, you change order of differentiation and integration and observe again, that the first part ($e^{\frac{-k_0x^2}{2kT}}$) again is constant and therefore not affected. This leaves you with just $e^\frac{-ax^4}{kT}$ to differentiate. This (again w.r.t. $a$) can be seen as $e^{a \cdot q}$ for some constant $q$, wich can very easyly be differetiated.