$\sum_{n=1}^\infty\int_0^\infty \left|ae^{-nax}-be^{-nbx}\right|dx=\infty$ I am trying to prove
$$\sum_{n=1}^\infty\int_0^\infty \left|ae^{-nax}-be^{-nbx}\right|dx=\infty$$
where $0<a<b$. I tried to bound it with $\sum_{n=1}^\infty\int_0^\infty ae^{-nax}-be^{-nbx}dx$ but that integral is equal to zero. I also tried to use
$$\int_0^\infty \sum_{n=1}^\infty\left|ae^{-nax}-be^{-nbx}\right|dx$$
but didn't know how. Any hint or suggestion?
 A: From the assumption $0<a<b$, it can be seen that $\frac{\ln(b)-\ln(a)}{(b-a)n}>0$ for every $n\geq 1$. With that in mind, let's split the integral into two as follows:
$$\int_0^\infty \left|ae^{-nax}-be^{-nbx}\right|dx=\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} \left|ae^{-nax}-be^{-nbx}\right|dx+\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty \left|ae^{-nax}-be^{-nbx}\right|dx$$
Notice that
\begin{align}
x\leq\frac{\ln(b)-\ln(a)}{(b-a)n} &\iff (b-a)nx\leq\ln(b)-\ln(a)\\
&\iff nbx-nax\leq \ln(b)-\ln(a)\\
&\iff \ln(a)-nax\leq\ln(b)-nbx\\
&\iff e^{\ln(a)-nax}\leq e^{\ln(b)-nbx}\\
&\iff ae^{-nax}\leq be^{-nbx}\\
&\iff ae^{-nax}-be^{-nbx}\leq0
\end{align}
This implies that
$$\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} \left|ae^{-nax}-be^{-nbx}\right|dx=-\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} \left(ae^{-nax}-be^{-nbx}\right)dx$$
Similarly
\begin{align}
x\geq\frac{\ln(b)-\ln(a)}{(b-a)n} &\iff (b-a)nx\geq\ln(b)-\ln(a)\\
&\iff nbx-nax\geq \ln(b)-\ln(a)\\
&\iff \ln(a)-nax\geq\ln(b)-nbx\\
&\iff e^{\ln(a)-nax}\geq e^{\ln(b)-nbx}\\
&\iff ae^{-nax}\geq be^{-nbx}\\
&\iff ae^{-nax}-be^{-nbx}\geq0
\end{align}
and thus,
$$\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty \left|ae^{-nax}-be^{-nbx}\right|dx=\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty \left(ae^{-nax}-be^{-nbx}\right)dx$$
We deduce that
\begin{align}
\int_0^\infty \left|ae^{-nax}-be^{-nbx}\right|dx &= \int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty \left(ae^{-nax}-be^{-nbx}\right)dx-\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} \left(ae^{-nax}-be^{-nbx}\right)dx\\
&= \int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty ae^{-nax}dx-\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty be^{-nbx}dx\\
&- \int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} ae^{-nax}dx+\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} be^{-nbx}dx\\
&= -\frac{1}{n}\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty -nae^{-nax}dx+\frac{1}{n}\int_\frac{\ln(b)-\ln(a)}{(b-a)n}^\infty -nbe^{-nbx}dx\\
&+ \frac{1}{n}\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} -nae^{-nax}dx-\frac{1}{n}\int_0^\frac{\ln(b)-\ln(a)}{(b-a)n} -nbe^{-nbx}dx\\
&= -\frac{1}{n}\int_{-na\left(\frac{\ln(b)-\ln(a)}{(b-a)n}\right)}^{-\infty}e^{u}du+\frac{1}{n}\int_{-nb\left(\frac{\ln(b)-\ln(a)}{(b-a)n}\right)}^{-\infty}e^{u}du\\
&+ \frac{1}{n}\int_0^{-na\left(\frac{\ln(b)-\ln(a)}{(b-a)n}\right)}e^{u}du-\frac{1}{n}\int_0^{-nb\left(\frac{\ln(b)-\ln(a)}{(b-a)n}\right)}e^{u}du\\
&= \frac{1}{n}\left(\int_{-\infty}^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-\infty}^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du+\int_{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du\right)\\
\end{align}
This shows that the terms of your series are constant multiples of $1/n$. Since the harmonic series $\sum_{n=1}^\infty 1/n$ diverges to $\infty$, your series will also diverge to $\infty$ if the constant prefactor, namely
$$\int_{-\infty}^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-\infty}^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du+\int_{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du$$
is positive. Now
\begin{align}
&\text{ }\int_{-\infty}^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-\infty}^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}e^{u}du-\int_{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du+\int_{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}^0 e^{u}du\\
&= e^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}-e^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}-\left(1-e^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}\right)+1-e^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}\\
&= e^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}-e^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}-1+e^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}+1-e^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}\\
&= 2e^{-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}-2e^{-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)}\\
\end{align}
and since the exponential function is one-to-one, this will be positive if and only if
$$-a\left(\frac{\ln(b)-\ln(a)}{b-a}\right)>-b\left(\frac{\ln(b)-\ln(a)}{b-a}\right)$$
This is true because $a<b$, so we’re done.
