# Thread: Does improper integral sin(x)/x converge?

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

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

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

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

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

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

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

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

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

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

so it converges to 0 by the squeeze theorem. So the limit exists, and therefore the integral converges.

2. Simplest solution:

Spoiler:

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

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