So, today we will talk about the conditional convergence and two discriminant methods, namely Dirac-Abel, which help us to decide whether a infinite integral is conditional convergence.
Definitions of absolute convergence and conditional convergence.
1. Absolute Convergence
$\displaystyle\int_{a}^{+\infty}f(x)dx$ and $\displaystyle\int_{a}^{+\infty}\left| f(x)\right|dx$ are both convergent.
By the way, the convergence of $\displaystyle\int_{a}^{+\infty}\left| f(x)\right|dx$ can actually deduce that $\displaystyle\int_{a}^{+\infty}f(x)dx$ is convergent.
2. Conditional Convergence
$\displaystyle\int_{a}^{+\infty}f(x)dx$ is convergent, but $\displaystyle\int_{a}^{+\infty}\left| f(x)\right|dx$ is not convergent.
Dirac-Abel Discriminant Methods(Dealing with Conditional Convergence).
1. Dirac Discriminant Method
if $\displaystyle\int_{a}^{x}f(u)du$ has the bound, and $\displaystyle g(x)$ is monotonic, $\displaystyle g(x)\to0$ when $\displaystyle x\to+\infty$,then
$\displaystyle\int_{a}^{+\infty}f(x)g(x)dx$ is convergent.
2. Abel Discriminant Method
if $\displaystyle\int_{a}^{+\infty}f(u)du$ is convergent, and $\displaystyle g(x)$ is monotonic and has the bound, then
$\displaystyle\int_{a}^{+\infty}f(x)g(x)dx$ is convergent.
Proof:
Before we prove these two discriminant methods, we need to first prove two related theorems, namely first and second mean value theorem for integral.
First mean value theorem for integral.
if $\displaystyle f(x)\in C[a,b]$, and $\displaystyle g(x)$ does not change the sign and is integrable in the $\displaystyle [a,b]$, then
$\displaystyle \int_{a}^{b}f(x)g(x)dx=f(\xi)\int_{a}^{b}g(x)dx$, in which $\xi$ is in the range of $[a,b]$.
Proof:
Since $\displaystyle f(x)\in C[a,b]$, then it must has minimum and maximum. Let's set them as $m$ and $M$, so
$\displaystyle m\leq f(x) \leq M$
Since $\displaystyle g(x)$ does not change the sign in the $\displaystyle[a,b]$, let's assume that $\displaystyle g(x)\ge0$. So, we multiply this inequality by $\displaystyle g(x)$ and get
$\displaystyle m g(x)\leq f(x)g(x) \leq M g(x)$
And we integral each element from $a$ to $b$, so
$\displaystyle m \int_{a}^{b}g(x)dx \leq \int_{a}^{b}f(x)g(x)dx \leq M \int_{a}^{b}g(x)dx$
If $\displaystyle \int_{a}^{b}g(x)dx = 0$, then the theorem is obviously correct.
If $\displaystyle \int_{a}^{b}g(x)dx \neq 0$, then since $\displaystyle g(x)\ge0$ in the $[a,b]$, we know that $\displaystyle \int_{a}^{b}g(x)dx > 0$, so we divide each element by $\displaystyle \int_{a}^{b}g(x)dx$, and get
$\displaystyle m\leq \frac{\int_{a}^{b}f(x)g(x)dx}{\int_{a}^{b}g(x)dx}\leq M$.
And since $\displaystyle f(x) \in C[a,b]$, according to intermediate value theorem, we get that
$\displaystyle f(\xi)=\frac{\int_{a}^{b}f(x)g(x)dx}{\int_{a}^{b}g(x)dx}$, in which $\xi$ is in the range of $[a,b]$.
Namely,
$\displaystyle \int_{a}^{b}f(x)g(x)dx=f(\xi)\int_{a}^{b}g(x)dx$, in which $\xi$ is in the range of $[a,b]$.
Second mean value theorem for integral.
if $\displaystyle f(x)\in C[a,b]$, and $\displaystyle g(x)$ is monotonic and differentiable in $[a,b]$, then
$\displaystyle \int_{a}^{b}f(x)g(x)dx = g(a)\int_{a}^{\xi}f(x)dx+g(b)\int_{\xi}^{b}f(x)dx$, in which $\xi$ is in the range of $[a,b]$.
Proof:
Set $\displaystyle F(x)=\int_{a}^{x}f(u)du\tag{$*$}$,then apply partial integeral, we get
$\displaystyle \int_{a}^{b}f(x)g(x)dx=F(x)g(x)\Big|_{a}^{b}-\int_{a}^{b}F(x)g'(x)dx$.
Namely,
$\displaystyle \int_{a}^{b}f(x)g(x)dx=F(b)g(b)-F(a)g(a)-\int_{a}^{b}F(x)g'(x)dx$.
Since $\displaystyle g(x)$ is monotonic, $\displaystyle g'(x)$ does not change sign in the $[a,b]$, then we apply the first mean value theorem for integral,
$\displaystyle F(b)g(b)-F(a)g(a)-\int_{a}^{b}F(x)g'(x)dx=F(b)g(b)-F(a)g(a)-F(\xi)\int_{a}^{b}g'(x)dx=F(b)g(b) - F(a)g(a)-F(\xi)(g(b)-g(a))$, in which $\xi$ is in the range of $[a,b]$.
So, by a few rearrangements,
$\displaystyle F(b)g(b)-F(a)g(a)-\int_{a}^{b}F(x)g'(x)dx=g(b)(F(b)-F(\xi))+g(a)(F(\xi)-F(a))$, in which $\xi$ is in the range of $[a,b]$..
Then, plug $(*)$ in(By the way, the integral variable does not matter in the difinite integral, so we can substitude $u$ with $x$),
$\displaystyle F(b)g(b)-F(a)g(a)-\int_{a}^{b}F(x)g'(x)dx = g(b)\int_{\xi}^{b}f(x)dx+g(a)\int_{a}^{\xi}f(x)dx$, in which $\xi$ is in the range of $[a,b]$.
Finally, we get
$\displaystyle \int_{a}^{b}f(x)g(x)dx = g(a)\int_{a}^{\xi}f(x)dx+g(b)\int_{\xi}^{b}f(x)dx$, in which $\xi$ is in the range of $[a,b]$.
Okay, and there is a last thing which we need to know to prove these two discriminant convergence. It's Cauchy's Convergence Test in the form of function. I will state it here but not prove it.
$\displaystyle \lim_{x\to +\infty}f(x)$ is convergent $\displaystyle \Leftrightarrow$ $\displaystyle \forall \epsilon > 0,\exists X > 0,\forall x_{1}>X,\forall x_{2}>X,\left|f(x_{1})-f(x_{2})\right|<\epsilon$.
Proof of Dirac Discriminant Convergence.
Based on the assumptions, set $\displaystyle \left|F(x)\right|=\left|\int_{a}^{x}f(u)du\right| \le M\tag{$\blacktriangle$}$, in which $\displaystyle x$ is in the range of $[a,+\infty)$ and $\displaystyle M > 0$.
And,
$\displaystyle \because g(x)$ is monotonic and goes to $0$ when $\displaystyle x \to +\infty$.
$\displaystyle \therefore \forall \bar{\epsilon}>0,\exists \bar{X}(\bar{\epsilon})>0,\forall x > \bar{X}, \left|g(x)\right|<\bar{\epsilon}\tag{$1$}$.
According to the difinition of infinite integral, $\displaystyle \int_{a}^{+\infty}f(x)g(x)dx \Longleftrightarrow \lim_{b\to +\infty}\int_{a}^{b}f(x)g(x)dx$.
If we want to prove,
$\displaystyle \lim_{b\to +\infty}\int_{a}^{b}f(x)g(x)dx$ is convergent.
based on the Cauchy's Convergence Test, we just need to prove that
$\displaystyle \forall \epsilon >0,\exists X>0,\forall x_{1}>X,x_{2}>X,\left|\int_{a}^{x_{2}}f(x)g(x)dx-\int_{a}^{x_{1}}f(x)g(x)dx\right| =\left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right|< \epsilon$.
So, for all $\epsilon > 0$,
In the $(1)$, we set $\displaystyle \bar{\epsilon}=\frac{\epsilon}{4M}$, and get $\displaystyle \exists \bar{X}(\bar{\epsilon})>0,\forall x > \bar{X}, \left|g(x)\right|<\bar{\epsilon}=\frac{\epsilon}{4M}\tag{$2$}$ (In the following text, $\bar{X}$ is refered to $\bar{X}(\bar{\epsilon})$)
for all $\displaystyle x_{1}>\bar{X}$ and $\displaystyle x_{2}>\bar{X}$,
Using the second mean value theorem for integral,
$\displaystyle \left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right| = \left|g(x_{1})\int_{x_{1}}^{\xi}f(x)dx+g(x_{2})\int_{\xi}^{x2}f(x)dx\right|\tag{$3$}$.
Using absolute value inequality,
$\displaystyle (3) \le \left|g(x_{1})\right|\left|\int_{x_{1}}^{\xi}f(x)dx\right|+\left|g(x_{2})\right|\left|\int_{\xi}^{x_{2}}f(x)dx\right|\tag{$4$}$
And that is,
$\displaystyle (4)=\left|g(x_{1})\right|\left|\int_{a}^{\xi}f(x)dx-\int_{a}^{x_{1}}f(x)dx\right|+\left|g(x_{2})\right|\left|\int_{a}^{x_{2}}f(x)dx-\int_{a}^{\xi}f(x)dx\right|$.
Using absolute value inequality again, and according to $(\blacktriangle)$ and $(2)$,
$\displaystyle (4) \le 2M(\left|g(x_{1})\right| + \left|g(x_{2})\right|) < 2M*2\bar{\epsilon}=\epsilon$.
Thus, by summing up, $\displaystyle \forall \epsilon >0,\exists X=\bar{X},\forall x_{1}>X,\forall x_{2}>X,\left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right| < \epsilon$, the theorem is proved.
Proof of Abel Discriminant Convergence.
The proof of Abel Discriminant Convergence is almost the same to the proof of Dirac Discriminant Convergence, so I will omit some trivial processes.
Based on the assumptions, let's set $\displaystyle \left|g(x)\right|\le M\tag{$\blacktriangle$}$, for $x$ in the range of $[a,+\infty)$, and in which $M > 0$.
$\displaystyle \because \int_{a}^{+\infty}f(x)dx$ is convergent
$\displaystyle \therefore \lim_{b \to +\infty}\int_{a}^{b}f(x)dx$ exists.
According to the Cauchy's Convergence Test,
$\displaystyle \forall \bar{\epsilon}>0,\exists \bar{X}(\bar{\epsilon})>0,\forall x_{1}>\bar{X},\forall x_{2}>\bar{X},\left|\int_{a}^{x_{1}}f(x)dx-\int_{a}^{x_{2}}f(x)dx\right|=\left|\int_{x_{1}}^{x_{2}}f(x)dx\right|<\bar{\epsilon}\tag{$1$}$
If we want to prove that $\displaystyle \lim_{b \to +\infty}\int_{a}^{b}f(x)g(x)dx$ is convergent, we just need to prove that
$\displaystyle \forall \epsilon >0,\exists X > 0,\forall x_{1}>X,\forall x_{2}>X,\left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right|<\epsilon$.
So, for all $\displaystyle \epsilon > 0$,
In $(1)$. let's set $\displaystyle \bar{\epsilon} = \frac{\epsilon}{2M}$, then $\displaystyle \exists \bar{X}(\bar{\epsilon}), \forall x_{1}>\bar{X},\forall x_{2}>\bar{X},\left|\int_{x_{1}}^{x_{2}}f(x)dx\right|<\bar{\epsilon}=\frac{\epsilon}{2M}\tag{$2$}$(In the following text, $\bar{X}$ is refered to $\bar{X}(\bar{\epsilon})$).
For $\displaystyle \forall x_{1}>\bar{X},\forall x_{2}>\bar{X}$, using the second mean value theorem for integral and absolute value inequality,
$\displaystyle \left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right|\le\left|g(x_{1})\right|\left|\int_{x_{1}}^{\xi}f(x)dx\right|+\left|g(x_{2})\right|\left|\int_{\xi}^{x_{2}}f(x)dx\right|\tag{$3$}$, in which $\xi$ is in the range of $[x_{1},x_{2}]$.
Combined with the $(\blacktriangle)$ and $(2)$(since $\xi > \bar{X}$ and notice $x_{1},x_{2}$ here are different from those in $(2)$),
$\displaystyle (3)\le 2M\bar{\epsilon}=\epsilon$
Thus, by summing up, $\displaystyle \forall \epsilon > 0,\exists X = \bar{X},\forall x_{1} > X,\forall x_{2}>X,\left|\int_{x_{1}}^{x_{2}}f(x)g(x)dx\right|<\epsilon$, the theorem is proved.