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I am currently doing some project and during the course of it I need to get an answer to the following:

Does $\displaystyle \int_e^\infty \frac{\sin x}{x \ln x}\;dx$ converge/ absolutely converge/diverge?

Thanks.

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The integral converges but not absolutely.

An elementary method is to consider, for every $n\geqslant3$, $$ I_n=(-1)^n\int\limits_{n\pi}^{(n+1)\pi}\frac{\sin x}{x\log x}\text{d}x, $$ to prove that the sequence $(I_n)_{n\geqslant3}$ is decreasing and converges to zero and to compare every integral $$ \int\limits_{\mathrm e}^{z}\frac{\sin x}{x\log x}\text{d}x=\int\limits_{\mathrm e}^{3}\frac{\sin x}{x\log x}\text{d}x+\int\limits_{3}^{z}\frac{\sin x}{x\log x}\text{d}x, $$ to a partial sum of the alternating series $\sum\limits_{n\geqslant3}(-1)^nI_n$.

The fact that the integral does not converge absolutely can be deduced from the estimation $I_n=\Theta(1/(n\log n))$ and from the divergence of the series $\sum\limits_{n\geqslant3}1/(n\log n)$.

Hints Let $n\geqslant3$ and $a_n=1/(n\log n)$.

(1) Since $|\sin|\leqslant1$, $I_n\leqslant\pi a_n$.

(2) Since $|\sin|\geqslant1/\sqrt2$ on an interval of length $\pi/2$ included in $(n\pi,(n+1)\pi)$ (which one?), $I_n\geqslant(\pi\sqrt2/4) a_{n+1}$.

(3) One can write each $I_n$ as the integral over $x$ in $(0,\pi)$ of the function $x\mapsto\sin(x)g_n(x)$ for a well chosen function $g_n$, and study the sequence $(g_n)_{n\geqslant3}$.

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    $\begingroup$ I'll upvote later, so for now I just wish to say that I like the technique of partitioning an infinite oscillatory integral into a sum of integrals over the zeroes of the oscillatory part. :) $\endgroup$ – J. M. is a poor mathematician Sep 4 '11 at 10:46
  • $\begingroup$ Absolutely agree $\endgroup$ – valdo Sep 4 '11 at 10:48
  • $\begingroup$ @J. M.: Just out of curiosity: if you like the answer, what's your reason for not upvoting right away? $\endgroup$ – Hans Lundmark Sep 4 '11 at 11:52
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    $\begingroup$ @Hans: I used up my upvotes four hours ago. $\endgroup$ – J. M. is a poor mathematician Sep 4 '11 at 11:53
  • $\begingroup$ @J. M.: Oh, I see! I didn't even know there was a limit... $\endgroup$ – Hans Lundmark Sep 4 '11 at 11:57
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The conditional convergence can also be handled using the integral version of the Dirichlet Test,

Suppose $f$ is a monotonic decreasing function so that $\lim\limits_{x\to\infty}f(x)=0$ and $g$ be an integrable function so that $\sup\limits_{x>a}\left|\int_a^xg(t)\;\mathrm{d}t\right|\le G<\infty$. Then, $\left|\int_a^\infty f(x)g(x)\;\mathrm{d}x\right|\le f(a)G$.

This is shown also using integration by parts: let $I(x)=\int_a^xg(x)\;\mathrm{d}x\le G$, then $$ \begin{align} \left|\int_a^\infty f(x)g(x)\;\mathrm{d}x\right| &\le\left|\lim_{x\to\infty}f(x)I(x)\right|+\left|f(a)I(a)\right|+\left|\int_a^\infty f'(x)I(x)\;\mathrm{d}c\right|\\ &\le0+0+\int_a^\infty -f'(x)G\;\mathrm{d}x\\ &=f(a)G \end{align} $$ QED

In this case, $f(x)=\dfrac{1}{x\log(x)}$ and $g(x)=\sin(x)$. Since $\int_a^x \sin(t)\mathrm{d}t=\cos(a)-\cos(x)$ and $\cos(x)\in[-1,1]$, the integral converges and its value is no greater than $2/e$.

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