Absolute convergence of $\int_0^{\infty}dy \int_0^{\infty} e^{-xy-1/y}\frac{\sin x}{1+x} dx$ The problem is to show absolute convergence of $I = \int_0^{\infty}dy \int_0^{\infty} e^{-xy-1/y}\frac{\sin x}{1+x} dx$.  I have already shown absolute convergence when the order of integration is swapped.  
Using $|\sin(x)| \le x$, $|\sin(x)| \le 1$, and $e^{-1/y} < 1$:
$$\int_0^{\infty}dy \int_0^1 \lvert e^{-xy-1/y}\frac{\sin x}{1+x}\rvert dx+ \int_1^{\infty} \lvert e^{-xy-1/y}\frac{\sin x}{1+x} dx \rvert \le$$
$$\int_0^{\infty}dy \int_0^1 e^{-xy}\frac{x}{1+x} dx+ \int_1^{\infty} \frac{e^{-xy-1/y}}{1+x} dx$$
Make the substitution $x = u + 1$ in the second integral:
$$ \int_1^{\infty} \frac{e^{-xy-1/y}}{1+x} dx = \int_0^{\infty} \frac{e^{-(u+1)y-1/y}}{2+u} du \le \int_0^{\infty} e^{-uy}e^{-y-1/y} du = \frac{e^{-y-1/y}}{y}$$
Make the substitution $x = 1/u$ in the first integral:
$$\int_0^1 e^{-xy}\frac{x}{1+x} dx = \int_1^{\infty} \frac{e^{-y/u}}{u^3+u^2} du \le \int_1^{\infty} \frac{e^{-y/u}}{u^3} du = $$
$$ \frac{1-e^{-y}(y+1)}{y^2}$$
We then have the resulting integral which can be shown to converge after a lot more work provided I didn't make any errors (which is quite possible since I have yet to check my work):
$$\int_0^{\infty} \frac{e^{-y-1/y}}{y} + \frac{1-e^{-y}(y+1)}{y^2} dy$$
But you get the picture that this is rapidly getting out of control.  I did spend much of the day trying to figure this problem out.  Is there a better way to do this?
 A: I should write the first integral as follows using $|\sin(x)| < x$ and $1+x \ge 1$:
$$\int_0^1 \lvert e^{-xy-1/y}\frac{\sin x}{1+x}\rvert dx \le \int_0^1 xe^{-xy-1/y} dx$$
Taking the derivative of the integrand:
$$ \frac{d}{dx}xe^{-xy-1/y} = e^{-xy-1/y}(1-xy)$$
we see that the integrand is $0$ at $x=0$ and increases to a maximum at $x = 1/y$ and then decreases back to $0$ as $x \rightarrow \infty$. Hence if $y \le 1$:
$$\int_0^1 xe^{-xy-1/y} dx \le \int_0^1 \frac{e^{-1-1/y}}{y} dx = \frac{e^{-1-1/y}}{y}$$
and if $y > 1$ then the maximum on the interval ${0 \le x \le 1}$ occurs at $x=1$:
$$\int_0^1 xe^{-xy-1/y} dx \le \int_0^1 {e^{-y-1/y}} dx = e^{-y-1/y}$$
Thus we have:
$$ \int_0^{\infty}dy \int_0^{\infty} |e^{-xy-1/y}\frac{\sin x}{1+x} dx| \le \int_0^{1} \frac{e^{-1-1/y}}{y} + \frac{e^{-y-1/y}}{y}dy + \int_1^{\infty} e^{-y-1/y} + \frac{e^{-y-1/y}}{y}dy \le$$
$$ 2\int_0^{1} \frac{e^{-1/y}}{y}dy + 2\int_1^{\infty} e^{-y}dy = 2*(\Gamma(0,1) + 1/e)$$
This is the simplest solution that I can come up with to show absolute convergence.  It took a long time to do it.  
