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Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral?

I am stuck with the following integral:

\begin{equation} \int_\mathbb{R} \frac{\sin t}{t} \end{equation}

I would like to find out whether this integral is convergent, but I totally forgot how to find the right convergence test here, at the origin I should have no problem since the integrand has a removable singularity. Towards infinity I was thinking about the harmonic series as a comparison, but obviously it is not going to bound the integrand from below, which would show divergence . Integration by parts should give me a logarithm multiplied by a trigonometric function, would that help? I then have to figure out $$\lim_{t \to \infty} \cos t \log t $$ .. how do I do this?

For the absolute integrand I reckon it is easier, here I can bound below using the harmonic series. But for the integral above I am not sure, thks for any hints!

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marked as duplicate by JavaMan, Srivatsan, Henning Makholm, Dylan Moreland, anon Feb 1 '12 at 7:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
I see two possible interpretations of the latter limit and they are incompatible; put some parentheses in it. (Not like the limit exists either way...) –  anon Feb 1 '12 at 2:17
    
This is done in Rudin's Real and Complex Analysis, second edition, on page 161, exercise 12. –  Michael Hardy Feb 1 '12 at 4:08
    
I wish I could accept all answers - thanks so much as well as for the comment of where to look at in Rudin's book !! I must have been blind yesterday wrt the limit posted above, (below the integral) of course $\cos t \log t$ increasingly oscillates as t gets large. Sometimes I have the eye of a mole. Anyways, many thanks !! –  harlekin Feb 1 '12 at 9:24

3 Answers 3

up vote 3 down vote accepted

Consider the "improper Riemman integral", \begin{align*} \int_0^\infty {\sin(x)\over x} \,dx &= \lim_{n\to\infty} \int_0^{\pi n}{\sin(x)\over x} \,dx \\ &= \lim_{n \to \infty}\sum_{k=0}^{n -1} \int_{k\pi}^{(k+1)\pi}{\sin(x)\over x}\,dx \\ &= \sum_{k = 0}^\infty (-1)^k \int_0^\pi{\sin(x)\over x + k\pi} \,dx \end{align*} The integrals are decreasing in $k$, so the series converges at $n\to\infty$ by the alternating series test. However, this integral is NOT absolutely integrable.

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I added in some limits and things that appeared to be missing. Hope I didn't mess it up! Nice answer. –  Dylan Moreland Feb 1 '12 at 2:42
    
I think you probably want $x+k\pi$ in the denominator of the last integral. –  robjohn Feb 1 '12 at 3:35
    
Thanks.... I get sleepy in the evening. –  ncmathsadist Feb 1 '12 at 11:13

A good idea to try whenever we have an integral with an oscillating factor and a dampening factor is to strengthen the dampening by integrating by parts. In this problem, behavior near $0$ is not the problem, so here it yields $$ \int^{\infty}_1 \frac{\sin x}{x} dx = \frac{- \cos x}{x} \biggr|^{\infty}_1 + \int^{\infty}_1 \frac{\cos x}{x^2} dx $$

which is certainly finite since the integral on the right hand side is absolutely convergent.

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This answer has been moved (and slightly modified) to an answer to a duplicate question.

The integral is not absolutely convergent. Because $\int_{k\pi}^{(k+1)\pi}|\sin(t)|\;\mathrm{d}t=2$, we have $$ \frac{2}{(k+1)\pi}\le\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin(t)}{t}\right|\;\mathrm{d}t\le\frac{2}{k\pi}\tag{1} $$ Since the harmonic series diverges, so does the integral of the absolute value.

However, the improper integral $$ \lim_{N\to\infty}\int_{-N}^N\frac{\sin(t)}{t}\mathrm{d}t\tag{2} $$ does exist. To see this, note that $$ \int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\mathrm{d}t=0\tag{3} $$ With $a=\frac12\left(\frac{1}{2k\pi}+\frac{1}{2(k+1)\pi}\right)$, recalling $\int_{2k\pi}^{2(k+1)\pi}|\sin(t)|\;\mathrm{d}t=4$, and using $(3)$, we get $$ \begin{align} \left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(t)}{t}\mathrm{d}t\right| &=\left|\int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\left(\frac1t-a\right)\;\mathrm{d}t\right|\\ &\le4\cdot\frac12\left(\frac{1}{2k\pi}-\frac{1}{2(k+1)\pi}\right)\\ &=\frac{1}{k(k+1)\pi}\tag{4} \end{align} $$ and $$ \sum_{k=1}^\infty\frac{1}{k(k+1)}=1\tag{5} $$ Therefore, $(1)$, $(4)$, and $(5)$ guarantee that $$ \int_{2\pi}^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{6} $$ converges to a value no greater than $\dfrac1\pi$.

Since $\left|\dfrac{\sin(t)}{t}\right|\le1$, $$ \int_0^{2\pi}\frac{\sin(t)}{t}\mathrm{d}t\tag{7} $$ has a value no greater than $2\pi$.

Since $\dfrac{\sin(t)}{t}$ is even, $(6)$ and $(7)$ guarantee that $$ \int_{-\infty}^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{8} $$ converges to a value no greater than $4\pi+\dfrac2\pi$.


Another general test is Dirichlet's test (Theorem 17.5). In this case, $$ \left|\int_0^N\sin(t)\;\mathrm{d}t\right|\le2 $$ and $\dfrac1t$ is monotonically decreasing to $0$ on $(0,\infty)$. Thus, by Dirichlet, $$ \int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t $$ converges.

By symmetry, $$ \int_{-\infty}^0\frac{\sin(t)}{t}\mathrm{d}t $$ converges also.

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