Asymptotics of integral

Question

I'm looking at the behavior of the following function $\Psi\left(\, z\, \right)$ around $0^{+}$ and $+\infty$, and I struggle a bit because of the integrability around $0\ldots$

$$ \Psi\left(\, z\, \right) = z^{1/2} \int_{0}^{\infty}\ln\left(1 + {\,\mathrm{e}^{-zy} \over y}\right) \,{\mathrm{d}y \over \,\sqrt{\,\vphantom{\large a} y\, }\,} + z^{3/2}\int_{0}^{\infty}\,\sqrt{\,\vphantom{\large a}y\, }\, \ln\left(1 + {\,\mathrm{e}^{-zy} \over y}\right)\,\mathrm{d}y . $$

If anyone could help me I'd be very happy !.

Answer 1

A possible hint for the FIRST integral (the second one could be quite similar).

$$\Psi_1(z) = \sqrt{z}\int_0^{+\infty} \frac{\ln\left(1 + \frac{e^{-zy}}{y}\right)}{\sqrt{y}}\ \text{d}y$$

Now use the substitution

$$zy = \ln t ~~~~~ \text{d}y = \frac{1}{zt}\ \text{d}t ~~~~~ t = e^{zy}$$

To get

$$\Psi_1(z) = \sqrt{z}\int_1^{+\infty}\ln\left(1 + \frac{e^{-\ln t}}{\frac{1}{z}\ln t}\right)\frac{\text{d}t}{zt} \frac{1}{\sqrt{\frac{1}{z}\ln t}}$$

Arranging a bit the terms to get:

$$\Psi_1 (z) = \int_1^{+\infty} \frac{1}{t}\ln\left(1 + \frac{z}{t\ln t}\right)\frac{\text{d}t}{\sqrt{\ln t}}$$

Now perfomr the other substitution

$$\ln t = p^2 ~~~~~ p = \sqrt{\ln t} ~~~~~\text{d}p = \frac{1}{2\sqrt{\ln t}}\frac{1}{t}\ \text{d}p$$

To get:

$$\Psi_1(z) = 2\int_0^{+\infty} \ln\left(1 + \frac{z}{p}e^{-p^2}\right)\ \text{d}p$$

I believe this is (maybe not) the best we can obtain from the first part. Maybe there are more suitable substitutions. I'm on the bus, so I hope I did not write malarkey.

One might try to expand the exponential term (like up to the second order) and give a try of the behaviour.

Otherwise, you might evaluate it with a $\Lambda$ cutoff like

$$\Psi_2(z) = \lim_{\Lambda\to \infty}\int_0^{\Lambda} \ln\left(1 + \frac{z}{p}e^{-p^2}\right)\ \text{d}p$$

For the second term, I'd try something similar.

Once at home, I'll take a deeper look.

Interesting case: $z = p$

$$\Psi_2(z = p) = \lim_{\Lambda\to \infty}\int_0^{\Lambda} \ln\left(1 + e^{-p^2}\right)\ \text{d}p = 0.6780938949040456$$

Answer 2

For $y \ll 1/z$, $e^{-zy}/y=1/y+O(1)$ so that $\ln(1+e^{-zy}/y)=\ln(e^{-zy}/y)+\ln(1+ye^{zy})$. Now $\ln(1/y+O(1))=-\ln(y)+O(y)$. Similarly $\ln(1+ye^{zy})=y+O(y^2)$. Thus $\ln(1+e^{-zy}/y)=-\ln(y)+O(y)$ in that limit.

For $y \gg 1/z$, $\ln(1+e^{-zy}/y)=e^{-zy}/y+O(e^{-2zy})$.

Thus your first integrand behaves like $-\ln(y)y^{-1/2}+O(y^{1/2})$ as $y \to 0$ and like $e^{-zy} y^{-3/2}+O(e^{-2zy})$ as $y \to \infty$. Your second integrand behaves like $-\ln(y) y^{1/2} + O(y^{3/2})$ as $y \to 0$ and like $e^{-zy} y^{-1/2}+O(e^{-2zy})$ as $y \to \infty$. So without diving into the world of composite asymptotic expansions (which can be quite complicated), an approximation can be given as

$$z^{1/2} \left ( \int_0^{a/z} -\ln(y) y^{-1/2} dy + \int_{a/z}^\infty e^{-zy} y^{-3/2} dy \right ) + z^{3/2} \left ( \int_0^{b/z} -\ln(y) y^{1/2} dy + \int_{b/z}^\infty e^{-zy} y^{-1/2} dy \right )$$

where $a,b$ are free parameters to be tuned for the approximation. Each of these integrals can be expressed in terms of the incomplete Gamma function (for the ones with the logarithms, you start with $u=\ln(y)$).