1
$\begingroup$

As a background I have proven that $\int_{[0,\infty[}e^{-xt}dt=\frac{1}{x}$

I am required to use Fubini's Theorem to calculate:

$\lim_{R\to \infty}\int_{[0,R]}\frac{\sin{x}}{x}d \lambda (x)$

I believe I know the manner in which to do this, but I believe I do not have the correct conditions to apply Fubini's theorem.

My ideas:

$\lim_{R\to \infty}\int_{[0,R]}\frac{\sin{x}}{x}d \lambda (x)=\lim_{R\to \infty}\int_{[0,R]}\sin{x}\int_{[0,\infty[}e^{-xt}d\lambda (t)d \lambda (x)$ $=\lim_{R\to \infty}\int_{[0,R]}\int_{[0,\infty[}\sin{x}e^{-xt}d\lambda (t)d \lambda (x)$

and then if I were able to apply Fubini:

$=\int_{[0,\infty[} \lim_{R\to \infty}\int_{[0,R]}\sin{x}e^{-xt}d\lambda (x)d \lambda (t)$

and since $[0,R]$ is a compact interval: Riemann-Integral = Lebesgue integral, therefore

$\int_{[0,\infty[} (\lim_{R\to \infty}\int_{[0,R]}\sin{x}e^{-xt}dx)d \lambda (t)$ etc.

as previously stated, my biggest problem is proving the condition necessary for Fubini's theorem, which would mean:

$\lim_{R\to \infty}\int_{[0,R]}\int_{[0,\infty[}|\sin{x}e^{-xt}|d\lambda (t)d \lambda (x)<\infty$, or $\lim_{R\to \infty}\int_{[0,R]}|\frac{\sin{x}}{x}|d \lambda (x)<\infty$

and I cannot prove any of those.

$\endgroup$
  • $\begingroup$ It is well known that $\int_{[0,\infty)} \frac{|\sin x|}{x} = +\infty$. This is the textbook example of a function that is not Lebesgue integrable, but the improper Riemann integral exists. $\endgroup$ – RRL Dec 17 '18 at 16:58
  • 1
    $\begingroup$ ... that is $\int_0^\infty \frac{\sin x }{x} \, dx $ converges but not absolutely. $\endgroup$ – RRL Dec 17 '18 at 17:08
2
$\begingroup$

Unfortunately the function $(x,t) \mapsto e^{-xt} \sin x$ is not absolutely integrable and Fubini's theorem does not apply. If it were then were then we could say

$$\int_0^\infty \int_0^\infty e^{-xt} | \sin x| \, dx \, dt = \int_0^\infty \int_0^\infty e^{-xt} | \sin x| \, dt \, dx $$

However,

$$\int_0^\infty e^{-xt} | \sin x| \, dt= \frac{|\sin x|}{x} $$

and this is neither Lebesgue nor improperly Riemann integrable over $[0,\infty]$.

You can justify

$$\int_0^\infty \int_0^\infty e^{-xt} \sin x \, dx \, dt = \int_0^\infty \int_0^\infty e^{-xt} \sin x \, dt \, dx $$

in terms of iterated (conditionally convergent) improper integrals by a combination of uniform and dominated convergence. See this answer for details.

$\endgroup$
2
$\begingroup$

Though $(x,t)\in (0,\infty)^2 \mapsto e^{-xt} \sin x$ is not absolutely integrable on the whole domain, we know that it is integrable on $[0,R]\times (0,\infty)$. Thus Fubini's theorem implies that for all $R>0$, it holds that $$\begin{eqnarray} \int_0^R\frac{\sin{x}}{x}d x &=& \int_0^R\sin{x}\int_0^\infty e^{-xt}dtdx\\ &=&\int_0^\infty\left(\int_0^R\sin{x} e^{-xt}dx\right)dt\\ &=&\int_0^\infty\frac{1-t\sin (R)e^{-Rt}-\cos (R)e^{-Rt}}{1+t^2}dt\\ &=&\frac{\pi}{2} - \int_0^\infty\frac{(t\sin (R)+\cos (R))e^{-Rt}}{1+t^2}dt. \end{eqnarray}$$ Note that the error term $$ |\int_0^\infty\frac{(t\sin (R)+\cos (R))e^{-Rt}}{1+t^2}dt|\leq C\int_0^\infty e^{-Rt}dt\to 0 $$as $R\to \infty.$ Thus by taking $R\to \infty$, we get $$ \lim_{R\to\infty}\int_0^R\frac{\sin{x}}{x}d x =\frac{\pi}{2}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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