Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove that $$g(x) = \text{p.v.} \int\limits_{-1/2}^{1/2}\frac{e^{-itx}}{t\cdot \ln{|t|}}dt $$ is function of $C_{0}(\mathbb{R})$. So, I need to prove that $$ \lim\limits_{|x|\to\infty}g(x)=0, $$ but I don't know how. Please help me.

share|improve this question
1  
What does "v.p." mean? –  Jonas Meyer May 22 '12 at 20:51
    
It is of help if you give us your definition of a principal value. What did you try? Where is your problem? What do you know? –  abatkai May 22 '12 at 20:54
2  
At least it is clear (Riemann-Lebesgue) that $$\int_{\varepsilon<|t|<1/2}\frac{e^{-itx}}{t\log |t|}dt\to0 \text{ as $|x|\to\infty$}$$ Also, $(\log|\log|t||)'=1/|t|\log t$. –  AD. May 22 '12 at 21:06
3  
@JonasMeyer I think it is an abbreviation for the French name valeur principale (Might it be from Cauchy?) –  AD. May 22 '12 at 21:31
1  
@JonasMeyer: I have usually seen it as $\text{p.v.}$ for "principal value" (and I have so corrected it since the OP has used $\text{p.v.}$ in another question). –  robjohn May 24 '12 at 0:42

1 Answer 1

up vote 6 down vote accepted

Continuing from where I left off in answer to another question, because $\frac{1}{t\log|t|}$ is odd, $$ \begin{align} \text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t &=-i\int_{-1/2}^{1/2}\frac{\sin(tx)}{t\log|t|}\,\mathrm{d}t\\ &=-2i\int_0^{1/2}\frac{\sin(tx)}{t\log|t|}\,\mathrm{d}t\\ &=-2i\int_0^{1/2}\frac{1}{t\log|t|}\,\mathrm{d}\frac{1-\cos(tx)}{x}\\ &=4i\frac{1-\cos(x/2)}{\log(2)x}+2i\int_0^{1/2}\frac{1-\cos(tx)}{x}\,\mathrm{d}\frac{1}{t\log|t|}\\ &=4i\frac{1-\cos(x/2)}{\log(2)x}-2i\int_0^{1/2}\frac{1-\cos(tx)}{x}\frac{1+\log(t)}{(t\log(t))^2}\,\mathrm{d}t\\ &=4i\frac{1-\cos(x/2)}{\log(2)x}-\frac{2i}{x}\int_0^{1/2}\frac{1-\cos(tx)}{t^2}\frac{1+\log(t)}{\log(t)^2}\,\mathrm{d}t\tag{1} \end{align} $$ We have that $(1)$ is odd and for $x>4$, $$ \left|4i\frac{1-\cos(x/2)}{\log(2)x}\right|\le\frac{8}{\log(2)x}\tag{2} $$ and $$ \begin{align} &\left|\frac{2i}{x}\int_0^{1/2}\frac{1-\cos(tx)}{t^2}\frac{1+\log(t)}{\log(t)^2}\,\mathrm{d}t\right|\\ &\le\frac2x\left(\int_0^{1/x}\frac{x^2}{2}\frac{1}{\log(x)}\mathrm{d}t+\int_{1/x}^{1/\sqrt{x}}\frac{2}{t^2}\frac{1}{\log(\sqrt{x})}\mathrm{d}t+\int_{1/\sqrt{x}}^{1/2}\frac{2}{t^2}\frac{1}{\log(2)}\mathrm{d}t\right)\\ &\le\frac2x\left(\frac{x}{2\log(x)}+\frac{4x}{\log(x)}+\frac{2\sqrt{x}}{\log(2)}\right)\\ &=\frac{9}{\log(x)}+\frac{4}{\log(2)\sqrt{x}}\tag{3} \end{align} $$ Thus, for $|x|>4$, $$ \left|\text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t\right|\le\frac{9}{\log|x|}+\frac{4}{\log(2)\sqrt{|x|}}+\frac{8}{\log(2)|x|}\tag{4} $$ So that $$ \lim_{|x|\to\infty}\text{p.v.}\int_{-1/2}^{1/2}\frac{e^{-itx}}{t\log|t|}\,\mathrm{d}t=0\tag{5} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.