Generalization of $\int_0^\infty \frac{e^{-cx} - e^{-dx}}{x} dx = \log(d/c)$ for $0I am trying to procure a generlization of the following result:
$\displaystyle \int_0^\infty \frac{e^{-cx} - e^{-dx}}{x} dx = \log(d/c)$ for $0<c<d$
This result is obtained by considering $$\int_a^b \int_c^d e^{-xy} dy dx = \int_c^d \int_a^b e^{-xy} dx dy$$
And the hint tells us to begin with
$$\int_a^b \int_c^d \varphi' (xy) dy dx$$
and introduce hypotheses in $\varphi$ where needed to justify the argument. 

The problem I have is that this problem is very vague (doesn't tell us what result we should get, and doesn't tell us what "suitable hypotheses" are), and I don't see a natural generalization of the result. However, this is what I have so far:
The first hypothesis I introduce is that $\frac{d}{dy} \varphi = \frac{d}{dx} \varphi$, so that I can switch integrals. 
So, I begin by integrating the inner integrals:
$$\int_a^b \varphi(dx) - \varphi(cx) dx = \int_c^d \varphi(ay) - \varphi(by) dy.$$
Let $a \to 0$ and $b \to \infty$. So I suppose that $\lim \limits_{a \to 0} \varphi(ay) = A$ and $\lim \limits_{b \to \infty} \varphi(by) = B$ for fixed, finite $y$. Now this becomes
$$\int_0^\infty \varphi(dx) - \varphi(cx) dx = \int_c^d (A - B) dy$$
but now I feel like I've gone off track because the result is different from the result that we're supposed to be generalizing.
 A: Begin with
$$\int_a^b \int_c^d \varphi' (xy) dy dx$$
Note that $\frac{1}{x}\frac{\partial \varphi}{\partial y}  = \frac{1}{y}\frac{\partial \varphi}{\partial x}$.  Thus, we have 
$$\begin{align}
\int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{x}\frac{\partial \varphi (xy)}{\partial y} dy dx \\
&=\int_a^b \frac{1}{x} \left( \varphi (dx)-\varphi (cx)\right) dx 
\end{align}$$
We also have
$$\begin{align}
\int_a^b \int_c^d \varphi' (xy) dy dx &=\int_a^b \int_c^d \frac{1}{y}\frac{\partial \varphi (xy)}{\partial x} dy dx \\
&=\int_c^d \frac{1}{y} \left( \varphi (by)-\varphi (ay)\right) dy 
\end{align}$$
Assume that $\lim_{z \to \infty} \varphi (z) =0$.  Then, as $a \to 0$ and $b \to \infty$, we see that
$$\begin{align}
\int_0^{\infty} \int_c^d \varphi' (xy) dy dx &=- \varphi (0) \int_c^d \frac{1}{y}  dx \\
&=-\varphi (0) \log (d/c)
\end{align}$$
Putting it all together, we have 
$$\int_0^{\infty} \frac{\left( \varphi (cx)-\varphi (dx)\right)}{x}  dx =\varphi (0) \log (d/c)$$
which generalizes the result!
