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Sine Integral — Definition, Formula & Examples

The sine integral, written Si(x), is a special function defined as the integral of sin(t)/t from 0 to x. It arises because the function sin(t)/t has no elementary antiderivative, so the integral itself is given a name and treated as a standalone function.

The sine integral is defined as Si⁡(x)=∫0xsin⁡tt dt\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}\, dt for all real xx. The integrand sinc⁡(t)=sin⁡tt\operatorname{sinc}(t) = \frac{\sin t}{t} is understood to equal 1 at t=0t = 0 by continuous extension. Si(x) is an odd, entire function with the limiting value Si⁡(∞)=π2\operatorname{Si}(\infty) = \frac{\pi}{2}.

Key Formula

Si⁡(x)=∫0xsin⁡tt dt=∑n=0∞(−1)n x2n+1(2n+1) (2n+1)!\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}\, dt = \sum_{n=0}^{\infty} \frac{(-1)^n\, x^{2n+1}}{(2n+1)\,(2n+1)!}
Where:
  • xx = Upper limit of integration; can be any real number
  • tt = Dummy variable of integration
  • nn = Index of summation (n = 0, 1, 2, ...)

How It Works

Because sin⁡tt\frac{\sin t}{t} cannot be integrated in terms of elementary functions, you evaluate Si(x) either by its Taylor series or numerically. The series representation is Si⁡(x)=∑n=0∞(−1)nx2n+1(2n+1)(2n+1)!\operatorname{Si}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)(2n+1)!}, obtained by integrating the Taylor series of sin⁡t\sin t term by term and dividing by tt. For small xx, a few terms of this series give excellent accuracy. For large xx, asymptotic expansions or numerical integration are used instead. The function increases from 0, overshoots π2\frac{\pi}{2}, then oscillates with decreasing amplitude, approaching π2\frac{\pi}{2} as x→∞x \to \infty.

Worked Example

Problem: Approximate Si(1) using the first three nonzero terms of the Taylor series.
Write the series terms: The first three terms of the series correspond to n = 0, 1, and 2.
Si⁡(1)≈11⋅1!−13⋅3!+15⋅5!\operatorname{Si}(1) \approx \frac{1}{1 \cdot 1!} - \frac{1}{3 \cdot 3!} + \frac{1}{5 \cdot 5!}
Evaluate each term: Compute the numerical value of each fraction.
=1−118+1600=1−0.05556+0.00167= 1 - \frac{1}{18} + \frac{1}{600} = 1 - 0.05556 + 0.00167
Sum the terms: Add the values together for the approximation.
Si⁡(1)≈0.94611\operatorname{Si}(1) \approx 0.94611
Answer: Si(1) ≈ 0.9461, which agrees with the known value 0.94608... to four decimal places.

Visualization

Why It Matters

The sine integral appears in signal processing when analyzing the Gibbs phenomenon near discontinuities in Fourier series. It also shows up in antenna theory and diffraction optics, making it a standard function in engineering reference tables alongside the cosine integral and exponential integral.

Common Mistakes

Mistake: Confusing Si(x) with the integral ∫0xsin⁡t dt=1−cos⁡x\int_0^x \sin t\, dt = 1 - \cos x.
Correction: Si(x) integrates sin(t)/t, not sin(t). The division by t is what makes the integral non-elementary and gives it special-function status.