5.5: Daunting Trigonometric Integral

The final example of taking out the big part is to estimate a daunting trigonometric integral that I learned as an undergraduate. My classmates and I spent many late nights in the physics library solving homework problems; the graduate students, doing the same for their courses, would regale us with their favorite mathematics and physics problems.

The integral appeared on the mathematical-preliminaries exam to enter the Landau Institute for Theoretical Physics in the former USSR. The problem is to evaluate

\[\int_{-\pi/2}^{\pi/2} (\cos t)^{100}dt \label{5.42} \]

to within 5% in less than 5 min without using a calculator or computer!

That \((\cos t)^{100}\) looks frightening. Most trigonometric identities do not help. The usually helpful identity \((\cos t)^{2} = (\cos 2t − 1)/2\) produces only

\[(\cos t)^{100} = (\frac{\cos 2t - 1}{2})^{50}, \label{5.43} \]

which becomes a trigonometric monster upon expanding the 50th power.

A clue pointing to a simpler method is that 5% accuracy is sufficient so, find the big part! The integrand is largest when \(t\) is near zero. There, \(\cos t ≈ 1 − t^{2}/2\) (Problem 5.20), so the integrand is roughly

\[(\cos t)^{100} ≈ (1 - \frac{t^{2}}{2})^{100}. \label{5.44} \]

It has the familiar form \((1 + z)^{n}\), with fractional change \(z = −t^{2}/2\) and exponent \(n = 100\). When \(t\) is small, \(z = −t^{2}/2\) is tiny, so \((1+z)^{n}\) may be approximated using the results of Section 5.3.4:

\[(1+z)^{n} \approx\left\{\begin{array}{ll}
1+n z & (z \ll 1 \text { and } n z \ll 1) \\
e^{n z} & (z \ll 1 \text { and } n z \text { unrestricted) }
\end{array}\right.\label{5.45} \]

Because the exponent \(n\) is large, \(nz\) can be large even when t and \(z\) are small. Therefore, the safest approximation is \((1 + z)^{n} ≈ e^{nz}\); then

\[(\cos t)^{100} ≈ (1 - \frac{t^{2}}{2})^{100} ≈ e^{-50t^{2}}. \label{5.46} \]

A cosine raised to a high power becomes a Gaussian! As a check on this surprising conclusion, computer- generated plots of \((cost)^{n}\) for \(n = 1...5\) show a Gaussian bell shape taking form as \(n\) increases.

Even with this graphical evidence, replacing \((cos t)^{100}\) by a Gaussian is a bit suspicious. In the original integral, t ranges from \(−\pi/2\) to \(\pi/2\), and these endpoints are far outside the region where \(\cos t ≈ 1 − t^{2}/2\) is an accurate approximation. Fortunately, this issue contributes only a tiny error (Problem 5.35). Ignoring this error turns the original integral into a Gaussian integral with finite limits:

\[\int_{-\pi/2}^{\pi/2} (cost)^{100}dt ≈ \int_{-\pi/2}^{\pi/2} e^{-50t^{2}}dt. \label{5.47} \]

Unfortunately, with finite limits the integral has no closed form. But extending the limits to infinity produces a closed form while contributing almost no error (Problem 5.36). The approximation chain is now

\[\int_{-\pi/2}^{\pi/2} (cos t)^{100}dt ≈ \int_{-\pi/2}^{\pi/2} e^{-50t^{2}}dt ≈ \int_{-\infty}^{\infty} e^{-50t^{2}}dt. \label{5.48} \]

Multiple problems

Problem 5.35 Using the original limits

The approximation \(\cos t ≈ 1 − t^{2}/2\) requires that t be small. Why doesn’t using the approximation outside the small-t range contribute a significant error?

Problem 5.36 Extending the limits

Why doesn’t extending the integration limits from ±\(\pi/2\) to ±\(\infty\) contribute a significant error?

The last integral is an old friend (Section 2.1): \(\int_{-\infty}^{\infty} e^{-at^{2}}dt = \sqrt{\pi/a}\). With \(α = 50\), the integral becomes \(\sqrt{π/50}\). Conveniently, 50 is roughly \(16\pi\), so the square root and our 5% estimate is roughly 0.25.

For comparison, the exact integral is (Problem 5.41)

\[\int_{-\infty}^{\infty} (cos t)^{n} dt = 2^{-n}(\left(\begin{array}{l} n \\ n/2 \end{array}\right): \pi \label{5.49} \]

When \(n = 100\), the binomial coefficient and power of two produce

\[\frac{12611418068195524166851562157}{158456325028528675187087900672}\pi ≈ 0.25003696348037. \label{5.50} \]

Our 5-minute, within-5% estimate of 0.25 is accurate to almost 0.01%!

Multiple problems

Problem 5.37 Sketching the approximations

Plot \((cos t)^{100}\) and its two approximations \(e^{-50t^{2}}\) and \(1 - 50t^{2}\).

Problem 5.38 Simplest approximation

Use the linear fractional-change approximation \((1 − t^{2}/2)^{100} ≈ 1 − 50t^{2}\) to approximate the integrand; then integrate it over the range where \(1 − 50t^{2}\) is positive. How close is the result of this 1-minute method to the exact value 0.2500 . . .?

Problem 5.39 Huge exponent

Estimate

\[\int_{-\pi/2}^{\pi/2} (cos t)^{100000}dt. \label{5.51} \]

Problem 5.40 How low can you go?

Investigate the accuracy of the approximation

\[\int_{-\pi/2}^{\pi/2} (cos t)^{n} dt ≈ \sqrt{\frac{\pi}{n}}, \label{5.52} \]

for small \(n\), including \(n = 1\).

Problem 5.41 Closed form

To evaluate the integral

\[\int_{-\pi/2}^{\pi/2} (\cos t)^{100} dt \label{5.53} \]

in closed form, use the following steps:

1. Replace \(\cos t\) with (\(e^{it} + e^{−it})/2\).
2. Use the binomial theorem to expand the 100th power.
3. Pair each term like \(e^{ikt}\) with a counterpart \(e^{−ikt}\); then integrate their sum from \(−\pi/2\) to \(\pi/2\). What value or values of \(k\) produce a sum whose integral is nonzero?