Gaussian type integral with negative power of variable in integrand How can we compute the integral $\int_{-\infty}^\infty t^n e^{-t^2/2} dt$ when $n=-1$ or $-2$? It is a problem (1.11) in Prof James Nearing's course Mathematical Tools for Physics. Can a situation arise in physics where this type of integral with negative power can be used?
 A: In this form, you won't get it naturally in physics because the Gaussian factor $\exp(-t^2)$ appears in the normal distribution or the harmonic oscillator etc. but with the addition of $1/t$ or $1/t^2$, one gets a non-normalizable and non-integrable (divergent) modification of the original Gaussian. However, one could surely engineer a situation in which the integral would arise in a different form.
The integral
$$\int_{-\infty}^\infty  t^n \exp(-t^2)$$
strictly speaking vanishes for $n$ odd because the integral is an odd function. Even more accurately, it is divergent for $n\leq -1$ and that's the strict answer to the OP's question. The behavior of the integrand near $t=0$ is simply $t^n$ whose indefinite integral is $t^{n+1}/(n+1)$ which is singular for $t=0$ and $n\leq -1$.
Less strictly, we may compute the analytically continued principal value – for an odd negative $n$, we have to add the absolute value to $t$, to get nonzero – and the integral
may be converted via substitution $t^2=T$ i.e. $t=\sqrt{T}$ and $dt=dT/2\sqrt{T}$ to 
$$2\int_{-\infty}^\infty T^{n/2-1/2} \exp(-T) dT/2 = (n/2-1/2)! $$
It's the Euler integral for the gamma function that I wrote as a generalized factorial. The factors of $2$ cancel. For $n=-1$, the argument of the factorial is $-1$ and the factorial itself is truly divergent – a pole. Note that it's exactly the point at which the strict principal value would vanish because of the odd nature of the factorial. For $n=-2$, the argument of the factorial is $-1.5$ and the result is
$$ (-1.5)! = (-0.5)! / 0.5 = 2\sqrt{\pi} $$
The strict integral would be divergent, even with the principal value, but the results above, $\infty$ and $2\sqrt\pi$, are likely those that would be obtained by a physicist who would "naturally" adjust the integral.
A: The integral may still exist in the sense of Analytic continuation. For example, we know that:
$$ \int_{-\infty}^{\infty}dx \; \exp(-ax^{2}) = \frac{ \sqrt \pi}{\sqrt a}$$
Now you can integrate inside the expression with respect to 'a' n times to get the integral
$$  \int_{-\infty}^{\infty}dx \; \exp(-ax^{2})\frac{(-1)^{n}}{x^{2n} }= C \frac{d^{-n}}{dx^{-n}}a^{-1/2}$$
