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.

  • $\begingroup$ 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

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

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


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

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.


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 $$ \lim_{R\to\infty}\int_0^R\frac{\sin{x}}{x}d x =\frac{\pi}{2}. $$


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