I have the following integrals : $$I_1(w)=\int_0^{+\infty} dy(1+\frac{1}{y+w^2})\frac{w^2\sqrt{y}}{e^y (y+w^2)-w^2}$$ $$I_2(w)=\int_0^{+\infty} dy(1+\frac{1}{y+w^2})\frac{w^2\sqrt{y}e^{-y}}{y+w^2}$$
Numerically, when $w \rightarrow 0$, I find that $$\lim_{w\rightarrow 0}\frac{I_1(w)}{I_2(w)}=2$$
I can't manage to prove it analytically. Any help would be very nice !
Make the change of variable $y = w^2 u$ in both integrals, to get
$$I_1 (w) = \int \limits _0 ^\infty \left( 1 + \frac 1 {w^2 u + w^2} \right) \frac {w^2 |w| \sqrt u} {\Bbb e ^{w^2 u} (u w^2 + w^2) - w^2} w^2 \ \Bbb d u = \\ |w| \int \limits _0 ^\infty \frac {w^2 u + w^2 + 1} {u + 1} \frac {\sqrt u} {\Bbb e ^{w^2 u} (u + 1) - 1} \ \Bbb d u $$
and
$$I_2 (w) = \int \limits _0 ^\infty \left( 1 + \frac 1 {w^2 u + w^2} \right) \frac {w^2 |w| \sqrt u} {u w^2 + w^2} w^2 \ \Bbb d u = |w| \int \limits _0 ^\infty \frac {w^2 u + w^2 + 1} {u + 1} \frac {\sqrt u} {u + 1} \ \Bbb d u .$$
When taking $w \to 0$, the only parts of those integrals that may contribute something non-zero are
$$J_1 (w) = |w| \int \limits _0 ^\infty \frac 1 {u + 1} \frac {\sqrt u} {\Bbb e ^{w^2 u} (u + 1) - 1} \ \Bbb d u$$
and, respectively,
$$J_2 (w) = |w| \int \limits _0 ^\infty \frac 1 {u + 1} \frac {\sqrt u} {u + 1} \ \Bbb d u $$
It follows that
$$\lim _{w \to 0} \frac {I_1 (w)} {I_2 (w)} = \lim _{w \to 0} \frac {J_1 (w)} {J_2 (w)}$$
so this gets rid of $|w|$ and of that little $\Bbb e ^{w^2 u}$ which becomes $1$, therefore your limits is precisely $\frac {K_1} {K_2}$ where
$$K_1 = \int \limits _0 ^\infty \frac 1 {u + 1} \frac {\sqrt u} {u} \ \Bbb d u = [u = t^2] = \int \limits _0 ^\infty \frac 2 {t^2 + 1} \ \Bbb d t = 2 \arctan t \Big| _0 ^\infty = \pi$$
and
$$K_2 = \int \limits _0 ^\infty \frac 1 {u + 1} \frac {\sqrt u} {u + 1} \ \Bbb d u = [u = t^2 ] = \int \limits _0 ^\infty \frac {2t^2} {(t^2 + 1)^2} = [t = \tan x] = \int \limits _0 ^{\frac \pi 2} 2 \sin^2 x \ \Bbb d x = \frac \pi 2 ,$$
where in the last integral I have used that $2 \sin^2 x = 1 - \cos 2x$.