Solution to $\int^{\infty}_0 \frac{1}{q} e^{-aq^2} dq.$

Question

My apologies if this question has already been asked. I've tried using the search function but could not find the answer.

I'm looking to solve the following integral:

$$ \int^\infty_0 \frac{1}{q} e^{-aq^2} dq, $$ where $a >0 $ and $q$ denotes the magnitude of the momentum. Now obviously this integral does not converge on the given domain but I am looking for find a way to extract a sensible approximation from this equation. The context is that I am calculating decoherence rates for various quantum mechanical interactions and when calculating the decoherence rate for møller scattering I am left with a bunch of constants and this integral. I would really appreciate any thoughts on this. Thank you in advance!

Edit_1: In response to the question of the wider physical context I will briefly outline the problem below.

I am calculating the rate at which several electromagnetic interactions cause a loss of entanglement in the spatial superposition of a charged particle in free-fall.

The particles are entangled with their environment such that their reduced density matrix becomes (not quite sure how the bra-ket notation works here as the braket latex package is not loading):

$$ \rho_{\mathcal{S}} = \sum_{n,m = 1}^{N} c_{n}c_{m}^{*} |\psi_{n}>< \psi_{m}| <E_{m}|E_{n}>. $$

Since the entanglement is generated using spin-coupling to a magnetic field we can express this using singlet spin states as:

$$ \rho_{\mathcal{S}} = \frac{1}{2} \begin{pmatrix} 1 & <E_{2}|E_{1}> \\ <E_{2}|E_{1}> & 1 \end{pmatrix}. $$

After several steps this gives us the time evolution of the off-diagonal terms as:

$$ \rho_\mathcal{S}(\textbf{x},\textbf{x}^{\prime},t) = \rho_\mathcal{S}(\textbf{x},\textbf{x}^{\prime},0)e^{-\gamma t} $$

This $\gamma$ is called the decoherence rate and is given by $\Gamma \equiv \int dq \varrho(q) v(q) \sigma_{tot}(q)$.

The calculation of this $\Gamma$ is giving me problems. Taking møller scattering (electron-electron scattering) as the relevant interaction we find that:

$$ \sigma \approx \frac{14\pi\alpha^2\hbar^2c^2m^2}{4\textbf{q}^4}, $$

$\varrho$ is given by the Maxwell-Boltzmann distribution:

$$ \varrho = \frac{N}{V} 4\pi q^2 \left( \frac{1}{2\pi m k_b T}\right)^{3/2} exp\left[-\frac{q^2}{2mk_bT}\right], $$

and $v = \frac{\textbf{q}}{m}$. This gives:

$$ \Gamma = \frac{N}{V} 14 m \pi^2 \alpha^2 \hbar^2 c^2 \left( \frac{1}{2\pi m k_b T2}\right)^{3/2} \int_0^\infty dq \frac{1}{q} exp\left[-\frac{q^2}{2mk_bT}\right]. $$

Where the final integral is the one which I am asking the question about. Since q is here is dependent on the thermal velocity (the environmental particles are in a gas) and the temperatures are low, on the order of 1-5K, we have that q is likely limited to not be extremely large. Perhaps this helps in understanding the problem.

Answer 1

Take $aq^2 = t$

Now, $dt$ = $2aq \ dq$

$\implies \frac{dq}{q} = \frac{dt}{2aq^2}$

$\implies \frac{dq}{q} = \frac{dt}{2t}$

Thus the integration is simplified to

$\int \frac{e^{-t}}{2t} \ dt$ which is an exponential integral.

Answer 2

As @Avnish Singh answered $$\int\frac{1}{q} e^{-aq^2}\, dq=\frac 12 \int \frac{e^{-t}}{ t}\,dt=\frac 12 \text{Ei}(-t)$$

Expanded as a series around $t=0$ gives for $$\text{Ei}(-t)=\log (t)+\gamma -t+\frac{t^2}{4}-\frac{t^3}{18}+O\left(t^4\right)$$ which is a very good approximation.

So $$\int_\epsilon^\infty\frac{1}{q} e^{-aq^2}\, dq=\frac{\log (a)+\gamma}{2}+\log(\epsilon)-\frac a 2 \epsilon^2+O\left(\epsilon^4\right)$$