I want to evaluate the divergent integral:

$$\int_0^{\infty} dx\; x^{-2} e^{-x^2}$$

My plan is to calculate the following integral instead,

$$ \int_0^{\infty} dx\; x^{-2g} e^{-x^2}= \Gamma\left(\frac{1}{2}-g\right) \qquad \text{if} \quad Re(g)<\frac{1}{2}. $$

Since $\Gamma\left(\frac{1}{2}-g\right)$ is analytic for $\frac{1}{2}<g < \frac{3}{2}$, I use $z\Gamma(z) = \Gamma(z+1)$, then

$$ \int_0^{\infty} dx\; x^{-2} e^{-x^2} = lim_{g \rightarrow 1} \frac{2 \;\Gamma\left(\frac{3}{2}-g\right)}{1-2g}= -2\Gamma(1/2) .$$

My question is, what conditions must be satisfied for the above calculations to be valid.


I am afraid that your calculation is never valid. The integral is not convergent at the lower bound. The definite integral from $x=0$ to $\infty$ is infinite.

In fact : $$\int x^{-2}e^{-x^2}dx =-\sqrt{\pi}^:\text{erf}(x)-\frac{e^{-x^2}}{x}+C\qquad \text{with}\qquad x\neq 0.$$

$$\int_\epsilon^\infty x^{-2}e^{-x^2}dx =-\sqrt{\pi}+\sqrt{\pi}^:\text{erf}(\epsilon)+\frac{1}{\epsilon}\qquad \text{with}\qquad \epsilon> 0.$$ Asymptotically : $$\int_\epsilon^\infty x^{-2}e^{-x^2}dx \sim \frac{1}{\epsilon}-\sqrt{\pi}\qquad \text{with}\qquad \epsilon\to 0.$$

If we remove the singularity at $x=0$, that is for example in considering the function $x^{-2}\left(e^{-x^2}-1\right)$ instead of $x^{-2}e^{-x^2}$ $$\int_0^\infty \frac{e^{-x^2}-1}{x^2}dx=-\sqrt{\pi}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.