Integration Techniques

Knowing that an option's fair price is the expected discounted payoff — an integral of the payoff against a probability density — is only half the job. You still have to actually compute that integral to get a number, and most integrals that show up in pricing don't have an obvious antiderivative sitting in front of you. Integration techniques are the systematic methods for turning an integral you can't immediately write down the answer to into one you can, by restructuring the expression algebraically first.

An analogy: unwrapping a nested package

Some integrals are like a small, single, plainly labeled box — you know the answer on sight. Others are like a package wrapped inside another wrapped package: you can't see what's inside until you undo the outer layer first. Integration techniques are the systematic ways of "unwrapping" a complicated integral — substitution repackages the problem in different, simpler variables; integration by parts pulls a product apart into a piece you can integrate directly plus a leftover piece that's hopefully easier than what you started with.

The math, one symbol at a time

Substitution reverses the chain rule: if you spot an inner function $g(x)$ and its derivative $g'(x)$ both present in the integrand, set $u = g(x)$, $du = g'(x)\,dx$, and rewrite the whole integral in terms of $u$:

$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.$$

In words: change variables so the "inner function and its own rate of change" collapse into a single simpler variable $u$, turning a tangled expression into a plain integral of $f$.

Integration by parts reverses the product rule:

$$\int u\,dv = uv - \int v\,du.$$

In words: split the integrand into two pieces, $u$ (which you'll differentiate, hopefully making it simpler) and $dv$ (which you'll integrate, hopefully staying manageable) — the original integral becomes a simple product $uv$ minus a new, hopefully easier, integral $\int v\,du$.

Worked example 1: substitution for a discounting integral

Compute $\int_0^{10} e^{-0.05t}\,dt$ (the total discount factor accumulated continuously over 10 years at a 5% rate). Let $u = -0.05t$, so $du = -0.05\,dt$, i.e. $dt = -20\,du$. The integral becomes $\int e^u \cdot(-20)\,du = -20e^u = -20e^{-0.05t}$, evaluated from 0 to 10: $-20e^{-0.5} - (-20e^0) = -20(0.6065) + 20 = -12.13 + 20 = 7.87$. So the integral equals approximately 7.87 — the substitution turned an integral with a messy exponent into one of the most basic integrals there is, $\int e^u\,du$.

Worked example 2: integration by parts for a duration-like integral

Compute $\int_0^{5} t\,e^{-0.1t}\,dt$ (relevant to computing a bond's Macaulay duration, which weights each cash flow's timing by its discounted size). Set $u = t$ (so $du = dt$) and $dv = e^{-0.1t}dt$ (so $v = -10e^{-0.1t}$). Then $\int u\,dv = uv - \int v\,du = -10t\,e^{-0.1t} - \int -10e^{-0.1t}\,dt = -10t\,e^{-0.1t} + 100(-e^{-0.1t}) \cdot(-1)$; carefully, $\int -10 e^{-0.1t}dt = 100e^{-0.1t}$, giving the antiderivative $-10te^{-0.1t} - 100e^{-0.1t}$. Evaluating from 0 to 5: at $t=5$, $-10(5)e^{-0.5} - 100e^{-0.5} = (-50-100)(0.6065) = -90.98$; at $t=0$: $0 - 100 = -100$. Result: $-90.98 - (-100) = 9.02$. The product-form integrand $t \cdot e^{-0.1t}$ — which had no obvious single antiderivative — was pulled apart by parts into pieces that each integrate cleanly.

What this means in practice

Continuous discounting, duration and convexity formulas, and the Black-Scholes derivation all involve integrals that need exactly these two techniques (or their probabilistic cousins, like the The Itô Integral in stochastic calculus) to reduce to closed form. Even where a pricing model is ultimately solved numerically, recognizing which piece of a formula is a standard integral — and simplifying it by hand first — often turns an expensive numerical integration into an exact, instant closed-form answer.