Asymptotic expansion of an integral

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.

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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

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.