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I got to prove that $$\int_0^1 \frac{1}{t}\sin\left(\frac{1}{t}\right)dt,$$ exists as an improper Riemann integral, yet that $$f(t)=\frac{1}{t}\sin\left(\frac{1}{t}\right)\notin \mathcal{L}_1((0,1),\mathbb{B},\lambda),$$ i.e. that $$\int_{[0,1]}\left|\frac{1}{t}\sin\left(\frac{1}{t}\right)\right|d\lambda=\infty.$$

Attempt: We know that $\forall \epsilon, 0<\epsilon \le 1$, $f(t)$ is continuous in $[\epsilon,1]$, so $\int_{\epsilon}^1 f(t)dt$ exists. Now, with the change of variable $z=1/t$ the integral becomes $$\int_1^{\frac{1}{\epsilon}} \frac{\sin(z)}{z}dz.$$ I have to prove the limit of the above expression converges as $\epsilon \rightarrow 0$.

For the second part I was told I should approximate $|f(t)|$ by simple functions and show that their integrals go to $\infty$, but I can't quite grasp the procedure.

Any ideas or insight would be greatly appreciated.

alonso s
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1 Answers1

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For the non integrability in the Lebesgue sense, one can use the intervals $$ I_n=\left(\dfrac1{n\pi+3\pi/4},\dfrac1{n\pi+\pi/4}\right). $$ On each $I_n$, $|\sin(1/t)|\geqslant1/\sqrt2$ and $t\leqslant1/(n\pi)$ hence $|\sin(1/t)/t|\geqslant n\pi/\sqrt2$. The length of $I_n$ is of order $1/n^2$ hence the integral of $|t\sin(1/t)|$ on $I_n$ is at least of order $1/n$. Summing these shows the Lebesgue integral diverges.

Proofs that $\displaystyle\int_1^x\frac{\sin(t)}t\mathrm dt$ converges when $x\to+\infty$ are all over the site.

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