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Math Help - Does improper integral sin(x)/x converge?

  1. #1
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    Does improper integral sin(x)/x converge?

    Does \int^\infty_0 \frac{\sin x}{x} \, dx converge?

    I'm having trouble understanding my professor's solution:

    \lim_{t \to 0^{+}} \int^\pi_t \frac{\sin x}{x} \,dx exists because \frac{\sin x}{x} extends to a continuous function at x = 0.

    Now this is where I'm having trouble. How are these equations derived?

    Let \lfloor x \rfloor be the floor function or greatest integer function. Then

    \lim_{t \to \infty} \int^{t}_\pi \frac{\sin x}{x} \, dx = \lim_{t \to \infty} \left( \sum_{n=2}^{\lfloor t/ \pi \rfloor} \int^{n \pi}_{(n-1)\pi} \frac{\sin x}{x} \, dx + \int^t_{\lfloor t/ \pi \rfloor \pi} \frac{\sin x}{x} \, dx \right)

    = \lim_{t \to \infty} \left( \sum_{n=2}^{\lfloor t/ \pi \rfloor} \int_0^\pi (-1)^{n-1} \frac{\sin x}{x + (n-1)\pi} \,dx + \int_{\lfloor t/ \pi \rfloor \pi}^t \frac{\sin x}{x} \,dx \right).

    Since \frac{\sin x}{x + (n-1)\pi} \le \frac{\sin x}{x + n \pi} for 0 \le x \le \pi, the first term has a finite limit as t \to \infty, by the Alternating Series Test. The second term is bounded in absolute value by

    \int_0^\pi \frac{\sin x}{x+ \lfloor t/\pi \rfloor \pi} \,dx \le \frac{1}{\lfloor t/\pi \rfloor},

    so it converges to 0 by the squeeze theorem. So the limit exists, and therefore the integral converges.
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  2. #2
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    Krizalid's Avatar
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    Simplest solution:

    Spoiler:

    Split the interval into two ones: \int_{0}^{1}{\frac{\sin x}{x}\,dx}+\int_{1}^{\infty }{\frac{\sin x}{x}\,dx}. First integral contains a removable discontinuity for the function at x=0, so the integral does exist, as for the second one, just integrate by parts and get \int_{1}^{\infty }{\frac{\sin x}{x}\,dx}=\int_{1}^{\infty }{\frac{(1-\cos x)'}{x}\,dx}=\left. \frac{1-\cos x}{x} \right|_{1}^{\infty }+\int_{1}^{\infty }{\frac{1-\cos x}{x^{2}}\,dx}, the last integral is non-negative, so direct comparison test applies and \frac{1-\cos x}{x^{2}}\le \frac{2}{x^{2}}, thus we're done.
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