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.

Here is an exercise from Dieudonné. He suggests to "perform integrations by part".

Let $f, g$ be positive $C^\infty$ functions, $F(x)=\int_1^x f(t)dt$ and assume that

  • $\int_1^\infty f(t) dt = +\infty$
  • $g'(t) \neq 0$ for $t \geq 1$, $\frac{f(x)}{g'(x)}$ is non-decreasing and goes to $+\infty$ with $x$.
  • $\frac{d}{dx} \left(\frac{f(x)}{g'(x)} \right) = f(x)h(x)$ where $h(x)=o(1)$ and $h'(x)=o(1)$ (so that $f(x) = o(F(x) g'(x))$ and $\log F(x) = o(g(x))$).

The question is: prove that $$ \int_1^x f(t) e^{i g(t)} dt \sim \frac{f(x)}{i g'(x)} e^{i g(x)} $$ when $x$ goes to infinity. I think I did all the possible integrations by parts, in vain.

I'll be grateful for any indication.

share|improve this question
    
It is not clear to me what the question is? –  Alice Nov 11 '11 at 15:13
    
Yes I've done this but I don't think it's obvious that the limit is 0. –  timofei Nov 13 '11 at 20:59

1 Answer 1

up vote 2 down vote accepted

Ok I got it. We want to show that $$ \int_1^x f(t) h(t) e^{i g(t)} dt \ll \frac{f(x)}{g'(x)} $$

Let's integrate by parts: $$ \int_1^x f(t) h(t) e^{i g(t)} dt = \left[ \frac{f(t) h(t)}{i g'(t)} e^{i g(t)} \right]_1^x - \frac{1}{i} \int_1^x \frac{d}{dt}\left( \frac{f(t)}{g'(t)} h(t) \right) e^{i g(t)} dt $$

and we have to handle the integral term on the right side. It it equal to $$ \int_1^x f(t) h(t)^2 e^{i g(t)} dt + \int_1^x \frac{f(t)}{g'(t)} h'(t) e^{i g(t)} dt $$

The integral on the left is $\ll \int_1^x f(t) e^{ig(t)} dt$ and we perform yet one more integration by parts on the right integral: $$ \int_1^x \frac{f(t}{g'(t)} h'(t) e^{i g(t)} dt = \left[ h(t) \frac{f(t)}{g'(t)} e^{i g(t)} \right]_1^x - \int_1^x h(t) \frac{d}{dt} \left( \frac{f(t)}{g'(t)} e^{i g(t)} \right) dt $$

Now it is easy, using the fact that $\frac{f}{g'}$ is increasing, to show that this last integral is $\ll \frac{f(x)}{g'(x)}$ and we are done.

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.