Prove that : 1. $\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$ 2. $\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$ Given the function sequence $f_n(x)=ae^{-anx}-be^{-bnx}$ where $0<a<b$ Prove that :

*

*$\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$

*$\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$
The second part I've managed to prove by just integrating and then it's a sum of zeroes($\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=\sum_{1}^{\infty}(-\frac{e^{anx}}{n}+\frac{e^{-bnx}}{n})|_{0}^{\infty}=\sum_{n=1}^{\infty}0=0$ but for the first part I've got no idea. any hint please
 A: Hint: For each $n$ find the unique $x_n>0$ such that
$$\tag 1 ae^{-anx_n} = be^{-bnx_n}.$$
On $(x_n,\infty)$ the left side of $(1)$ is greater than the right side. Thus
$$\int_0^\infty|f_n| > \int_{x_n}^\infty|f_n| =\int_{x_n}^\infty f_n.$$
A: find out when $f_n>0$ or $<0$ so:
$$0=f_n=ae^{-anx}-be^{-bnx}$$
$$a(e^{-nx})^a=b(e^{-nx})^b$$
$$\frac ab=(e^{-nx})^{b-a}$$
$$\ln a-\ln b=(b-a)(-nx)$$
so:
$$x=\frac{\ln a-\ln b}{n(a-b)}$$
is when $f_n=0$

$$\int_0^\infty f_ndx=\int_0^\infty\left(ae^{-anx}-be^{-bnx}\right)dx=\left[\frac{e^{-anx}-e^{-bnx}}{-n}\right]_{x=0}^\infty=0$$

$$\int_0^\infty|f_n|dx=-\int_0^{\frac{\ln a-\ln b}{n(a-b)}}(ae^{-anx}-be^{-bnx})dx+\int_{\frac{\ln a-\ln b}{n(a-b)}}^\infty(ae^{-anx}-be^{-bnx})dx$$
Try and find the value of this then plug it into the summation
A: Let $n\in \mathbb{N}$ fixed, now note that  $$a\exp(-ax)=b\exp(-bx) \iff x={{\ln\left(\frac{b}{a}\right)}\over {b-a}}$$Call this positive number $t$, then for $x\in [0,t/n]$ we have that $$b\exp(-bnx)\ge a\exp(-anx)$$
Therefore, e have \begin{eqnarray*}
\int_0^\infty |a\exp(-anx)-b\exp(-bnx)|dx &\ge& \int_0^{t/n} (b\exp(-bnx)-a\exp(-anx))dx\\
&=&{1\over n} \int_0^t(be^{-bt}-ae^{-at})dt
\end{eqnarray*} This integral is constant and positive, and the result follows
