Integrals & Estimation Tricks (IPhO Math Tools)

IPhO math tools lesson on integrals: substitutions that pull out parameters, symmetry and parts, quick standard integrals, and sharp estimation methods when exact answers are unnecessary.

  • International Physics Olympiad preparation
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Many IPhO integrals look scary until you do one smart move: a substitution that exposes the scale, a symmetry argument, or a recognition that you only need an estimate. This lesson focuses on the moves that most often convert “hard” integrals into one-line results.

1. Definitions (Must Know)

A. Definite vs indefinite integrals

  • Indefinite integral: ∫ f(x) dx = F(x) + C (an antiderivative).
  • Definite integral: ∫ₐ^b f(x) dx (a number, often an area or accumulated quantity).

B. Substitution (change of variables)

If u = u(x), then

∫ f(u) du = ∫ f(u(x))du/dx dx.

Always change limits for definite integrals.

C. Integration by parts

From the product rule:

∫ u dv = uv - ∫ v du.

Good for: polynomials times exponentials/trig, and for turning an integral into something simpler or bounded.

D. Improper integrals

Integrals with infinite limits or singularities are defined by limits, for example:

∫₀^∞ f(x) dx = lim _(R → ∞)∫₀^R f(x) dx.

E. Estimation language

If an integrand has typical size A over a width w, then a first estimate is

∫ f(x) dx ∼ Aw.

This becomes accurate when the integrand is sharply peaked (or nearly constant) in the region that contributes most of the area.

2. Key Ideas (What Earns Marks)

  • Use a scaling substitution to pull parameters outside the integral.
  • Look for symmetry (even and odd functions, periodicity, geometric symmetry).
  • If you see a product, consider integration by parts.
  • If the integral depends on a parameter, consider differentiating with respect to that parameter.
  • When exact evaluation is unnecessary, bound and estimate:
    • identify where the integrand is large,
    • approximate locally (often by a Gaussian),
    • state the expected error scale.

3. Detailed Explanations

A. Scaling substitution (parameter extraction)

If an integral contains a parameter a only through x/a, try x = a u.

Example pattern:

∫₀^∞ e^(-x/a) dx = a∫₀^∞ e^(-u) du ∼ a.

The physics interpretation: the integrand decays on a length scale set by a.

B. Symmetry (even, odd, and geometry)

If f(-x) = -f(x) (odd), then

∫_(-A)^A f(x) dx = 0.

If f(-x) = f(x) (even), then

∫_(-A)^A f(x) dx = 2∫₀^A f(x) dx.

These are time-saving when integrals arise from center-of-mass, field, or average-value calculations.

C. Parts as a “reduction move”

Integration by parts is often used to:

  • reduce powers (for example, ∫ xⁿ e^(-x) dx),
  • move derivatives onto an easier factor,
  • generate recursion relations (useful for ∫₀^(π/2) sinⁿ θ dθ).

D. Differentiate under the integral sign (Feynman trick)

If

I(α) = ∫ₐ^b f(x,α) dx,

and the integral behaves well, then

dI/dα = ∫ₐ^b (∂ f)/(∂ α) dx.

You differentiate to get a simpler integral, then integrate back in α using an easy boundary value.

E. Laplace method (sharp peak estimates)

If f(x) has a maximum at x₀ and the integral contains e^(n f(x)) with large n, the main contribution comes from near x₀.

Expand:

f(x) ≈ f(x₀) + 1/2 f''(x₀)(x-x₀)²

and the integral often becomes approximately Gaussian.

You rarely need full rigor in IPhO scripts, just a clear dominant-region argument.

4. Common Mistakes

  • Doing a substitution but forgetting the dx Jacobian or the new limits.
  • Losing a minus sign when reversing limits.
  • Integrating by parts in a loop (you end up with the same integral again without progress).
  • Treating an improper integral as convergent without checking decay at infinity or behaviour at a singular point.
  • Estimating without stating where the integral “gets its area”.

5. Exam Tips

  • Decide early: exact integral, or estimate. If the question asks for scaling, do not over-integrate.
  • If you introduce a substitution, write it and the transformed limits explicitly (one line each).
  • Keep a mini mental list of standard results:
    • Gaussian integrals
    • ∫₀^∞ e^(-ax) dx
    • ∫ dx/(a² + x²) = 1/a arctan(x/a)
  • For estimates, write “dominant region” in words. This often earns method credit.

6. Worked Examples

A. A Gaussian normalization integral

Evaluate ∫_(-∞)^∞ exp(-x²/2σ²) dx.

Click here to show/hide solution

Let

I = ∫_(-∞)^∞ exp(-x²/2σ²) dx.

Scale out σ with x = σ u:

I = σ∫_(-∞)^∞e^(-u²/2) du = σ J.

Compute J by squaring:

J² = (∫_(-∞)^∞e^(-u²/2) du)(∫_(-∞)^∞e^(-v²/2) dv) = ∬_R²e^(-(u² + v²)/2) dudv.

Convert to polar coordinates (r,θ) where u² + v² = r² and dudv = r dr dθ:

J² = ∫₀^2π∫₀^∞ e^(-r²/2) r dr dθ = 2π[-e^(-r²/2)]₀^∞ = 2π.

So J = square root of 2π and therefore

I = σ square root of 2π .

B. A rational integral that looks harder than it is

Evaluate ∫₀^Rr²/(a² + r²) dr for a > 0.

Click here to show/hide solution

Rewrite the integrand:

r²/(a² + r²) = 1-a²/(a² + r²).

Then:

∫₀^Rr²/(a² + r²) dr = ∫₀^R1 dr - a²∫₀^Rdr/(a² + r²) = R-a² · 1/a arctan(R/a).

So:

∫₀^Rr²/(a² + r²) dr = R-a arctan(R/a).

C. A standard improper integral with a fast complex method

For a > 0, evaluate ∫₀^∞e^(-ax) cos(bx) dx.

Click here to show/hide solution

Use cos(bx) = Re(e^ibx):

∫₀^∞e^(-ax) cos(bx) dx = Re(∫₀^∞ e^(-(a-ib)x) dx).

Since a > 0:

∫₀^∞ e^(-(a-ib)x) dx = 1/(a-ib).

Take the real part:

Re(1/(a-ib)) = Re((a + ib)/(a² + b²)) = a/(a² + b²).

Therefore:

∫₀^∞e^(-ax) cos(bx) dx = a/(a² + b²).

7. Mind Stretchers

A. Estimating ∫₀^(π/2) sinⁿ θ dθ for large n

Let

Iₙ = ∫₀^(π/2) sinⁿ θ dθ.

Estimate Iₙ for n≫ 1.

Click here to show/hide hint

For large n, the integrand is sharply peaked near θ = π/2.

Let u = π/2-θ with u small. Then sin θ = cos u ≈ 1-u²/2 and

sinⁿ θ ≈ exp(n ln(1-u²/2)) ≈ e^(-nu²/2).

Extend the upper limit to infinity.

Click here to show/hide answer

Let u = π/2-θ. For the region that matters when n≫ 1, u is small, so:

sin θ = cos u ≈ 1-u²/2.

Then:

ln(sinⁿ θ) = n ln(1-u²/2) ≈ -(n u²)/2,

so

sinⁿ θ ≈ e^(-n u²/2).

Thus:

Iₙ = ∫₀^(π/2) sinⁿ θ dθ ≈ ∫₀^∞ e^(-n u²/2) du.

Scale u = square root of (2/n) y:

∫₀^∞ e^(-n u²/2) du = square root of (2/n) ∫₀^∞e^(-y²) dy = square root of (2/n) · (square root of π)/2 = square root of (π/2n) .

So Iₙ ∼ square root of (π/2n) for large n.

8. Practice

Practice

Train “one smart move” integrals:

  • practice 10 substitutions that pull parameters outside
  • redo the Gaussian integral derivation once (it comes up everywhere)
  • for estimation, always state the dominant region and the scale of the width
Syllabus and review details

No official syllabus alignment is listed for this lesson.