Hello,
I need to show that an integral from minus infinity to infinity is smaller than infinity. All the expressions that are integrated are positive. What should I do? Can you recommend me some theorems that I can use? Thank you in advance!
Hello,
I need to show that an integral from minus infinity to infinity is smaller than infinity. All the expressions that are integrated are positive. What should I do? Can you recommend me some theorems that I can use? Thank you in advance!
$\displaystyle \int_{-\infty}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$
As |x| and x^2 are positive, we can write this as,
$\displaystyle 2\int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$
So checking the convergence of $\displaystyle \int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$ is enough.
$\displaystyle \int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx = \left [ \int_{0}^{1} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx \right ] ~+~ \left [ \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx \right ]$
We can easily prove that $\displaystyle \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$ converges.
For $\displaystyle 1\leq x \leq \infty$,
$\displaystyle 0\leq \frac{\ln x \cdot a \cdot e^{-\frac{a}{x^2}}}{x^3} \leq \frac{\ln x \cdot a}{x^3}$
$\displaystyle \int_1^{\infty} \frac{\ln x \cdot a}{x^3}~dx = \frac{a}{4}$
So $\displaystyle \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$ converges too.
Now prove the convergence of $\displaystyle \int_{0}^{1} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx$ and it'll be over.
(It looks a little long and I can't think of anything easy at the moment..)