Graduate Proving that this integral is divergent

[cbarker1]

TL;DR

I have a particular integral that I need to show that it cannot be finite in L1(positive real line) space. The integral comes from an Euler-Cauchy ODE. I want to show that the answer is not bounded.

Dear everyone,

I have a question on how to show that an integral is divigent. Here is the setup:
Suppose that we have the following function $\sigma(x)=\frac{1}{x^{2-\varepsilon}}$ for an arbitrary fixed $\varepsilon>0.$
\begin{equation}
\dfrac{d}{dx}[-u'(x)]=\dfrac{1}{x^{2-\varepsilon}}u
\end{equation}
We want to verify using the bounds and using explicit form of the solution $u$.

First we can rewrite the equation:
\begin{equation}
-u''(x)=u(x)\dfrac{1}{x^{2-\varepsilon}}
\end{equation}
Then we define a solution form of the differential equation $u=x^{r-\varepsilon}$ such that $u$ is in $C^{2}(0,\infty)$, where $C^{2}(0,\infty)$ is the set of all continuously differentiable functions on $(0,\infty).$ Then we know that $u''=(r-\varepsilon)(r-\varepsilon-1)x^{r-\varepsilon-2}$. Thus we have the following differential equation:
\begin{equation}
(r-\varepsilon)(r-\varepsilon-1)x^{r-\varepsilon-2}\cdotp x^{2-\varepsilon}+x^{r-\varepsilon}=0.
\end{equation}
Since #\varepsilon# is arbitrary, we can simplify the equation and factor $x^{r-\varepsilon}$ for $x^{r-\varepsilon}>0;$ therefore, we have the simplified equation:
\begin{equation}
(r-\varepsilon)(r-\varepsilon-1)+1=0.
\end{equation}
Now we must solve for $r$ in the auxiliary equation. We will expand the left-hand side and use the quadratic formula for $r$:
\begin{equation}
r=\dfrac{2\varepsilon+1\pm \sqrt{(2\varepsilon+1)^2-4\bigg(\dfrac{(2\varepsilon+1)^2+3}{4}\bigg)}}{2}
\end{equation}
After some simplifications on the previous equations, we have the following solution:
$r=\dfrac{2\varepsilon+1\pm i\sqrt{3}}{2}.$
Now the solution to the differential equation is
$u(x)=x^{r-\varepsilon}$ which implies that $u(x)=c_1\sqrt{x}\cos(\frac{\sqrt{3}}{2}\log x)+c_2\sqrt{x}\sin(\frac{\sqrt{3}}{2}\log x)$. So we apply the one of the boundary condition that $u(0)=0$; thus, the final solution is $u(x)=\sqrt{x}\cos(\frac{\sqrt{3}}{2}\log x).$ Now we need to check if the solution is not bounded.
In other words,
\begin{equation}
\sqrt{x}\cos\bigg(\frac{\sqrt{3}}{2}\log x\bigg)\geq \bigg(\int_{0}^{x}t^{-\varepsilon}dt\bigg)\bigg(\int_{x}^{\infty}t^{-\frac{3}{2}+\varepsilon}\cos(\frac{\sqrt{3}}{2}\log(t))dt\bigg)\\
\end{equation}
The red second integral is where I am having the most trouble on showing that it is not bounded.
Any suggestions are helpful. I thought to use a contradiction and squeeze theorem to show that the integral is not bounded.

Thanks,
Cbarker1

 

[anuttarasammyak]

t:= log x, the given quantity, say f(t), is
e^{\frac{t}{2}} cos( \frac{\sqrt{3}}{2}t)
It would make an estimation easier.
For any given number $A>\frac{4\pi}{\sqrt{3}} =7.25...$ there exists number a such that $A<a<2A,\ |cos( \frac{\sqrt{3}}{2}a)|=1.$ so
|f(a)|=e^{\frac{a}{2}} &gt; e^{\frac{A}{2}} &gt; A
unbounded.

 

[pasmith]

... the final solution is $u(x)=\sqrt{x}\cos(\frac{\sqrt{3}}{2}\log x).$ Now we need to check if the solution is not bounded.

Take a sequence x_n where \cos(x_n) = 1; solving \frac{\sqrt{3}}{2}\log x = 2n\pi gives x_n = \exp\left( \frac{4n\pi}{\sqrt{3}} \right). Now <br /> u(x_n) = \sqrt{x_n}\cos(2n\pi) = \exp\left( \frac{2n \pi}{\sqrt{3}} \right) increases without bound as n \to \infty.

In general, if you want to find how g(x) \cos f(x) or g(x) \sin f(x) behave as x \to \infty, look at those values of x for which f(x) = n\pi (for cos) or (n + \frac12)\pi (for sin). These provide bounds for the extrema of the oscillation.