Suppose $f$ is integrable. Prove: $\lim_{\lambda \to\infty} \int_{-a}^a \frac{1-\cos(\lambda x)}{x}f(x) dx=\int_0^a \frac{f(x)-f(-x)}{x}dx$

Question

Suppose $f$ is integrable. Prove: $\lim_{\lambda \to\infty} \int_{-a}^a \frac{1-\cos(\lambda x)}{x}f(x) dx=\int_0^a \frac{f(x)-f(-x)}{x}dx$

I got this problem from a friend, so I do not know the exact assumptions, but for this sake lets just assume that $f$ is continuous and differentiable on $[-a,a]$.

Attempt:$\lim_{\lambda \to\infty} \int_{-a}^a \frac{1-\cos(\lambda x)}{x}f(x) dx=\int_0^a \frac{f(x)-f(-x)}{x} dx +\lim_{\lambda\to\infty} \int_{-a}^{a} \frac{f(x)\cos(\lambda x)}{x} dx$

Now, I just need to show that $\lim_{\lambda\to\infty} \int_{-a}^{a} \frac{f(x)\cos(\lambda x)}{x} dx=0$

I am pretty sure the Riemann-Lesbesgue lemma should be helpful here, but alas I cannot see why $\frac{f(x)}{x}$ is integrable on $[-a,a]$ in order to apply it.

Thanks!

Answer 1

We can write $$ \int_{ - a}^a {\frac{{1 - \cos (\lambda x)}}{x}f(x)dx} = \int_0^a {\frac{{f(x) - f( - x)}}{x}dx} - \int_0^a {\cos (\lambda x)\frac{{f(x) - f( - x)}}{x}dx} . $$ Since $$ \mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f( - x)}}{x} = 2f'(0), $$ we can apply the Riemann$-$Lebesgue lemma and arrive at the desired result.

Answer 2

Split $f$ into its odd and even parts, defined by

$$f_o(x) \equiv \frac{f(x)-f(-x)}{2} \hspace{24 pt} f_e(x) \equiv \frac{f(x)-f(-x)}{2}$$

which has the property that $f(x) = f_e(x) + f_o(x)$. Now in the integral we have that

$$\int_{-a}^a \frac{1-\cos(\lambda x)}{x}f(x)\:dx = \int_{-a}^a \frac{1-\cos(\lambda x)}{x}\left[f_e(x)+f_o(x)\right]\:dx$$

Then since the other part of the integrand is an odd function, only the integral with the odd part of $f$ survives and we can use a symmetry argument to claim that

$$\int_{-a}^a \frac{1-\cos(\lambda x)}{x}f_o(x)\:dx = \int_0^a [f(x)-f(-x)]\frac{1-\cos(\lambda x)}{x}\:dx$$

Since $f_o(0) = 0$, we can move the $x$ over to the $f$ and get

$$ = \int_0^a \frac{f(x)-f(-x)}{x}\:dx - \int_0^a \frac{f(x)-f(-x)}{x}\cos(\lambda x)\:dx$$

For the second integral, use integration by parts

$$ = \frac{f(a)-f(-a)}{a}\frac{\sin(\lambda a)}{\lambda} - \frac{1}{\lambda}\int_0^a \left(\frac{f(x)-f(-x)}{x}\right)'\sin(\lambda x)\:dx$$

which goes to $0$ by squeeze theorem when taking the limit, leaving us with our final result:

$$\lim_{\lambda \to \infty} \int_{-a}^a \frac{1-\cos(\lambda x)}{x}f(x)\:dx = \int_0^a \frac{f(x)-f(-x)}{x}\:dx$$