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Integration Techniques & Improper Integrals

Useful preparation: Integrals & the Fundamental Theorem

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Meaning & notation

Integration methods reverse derivative patterns: substitution follows composition, parts rearranges a product rule, and partial fractions simplify rational expressions. For improper integrals, also ask whether finite-cutoff values approach a finite limit.

Substitution: Replacing an inner expression by a new variable while transforming its differential and, for a definite integral, its endpoints consistently.
Integration by parts: Trading an integral of one product for another using the product rule. A useful choice makes the remaining integral easier to evaluate.
Improper integral: An integral defined through one-sided limits because an interval is unbounded or the integrand is unbounded near an endpoint or interior point.

Conditions before calculation

Use substitution on intervals where the chain-rule identity applies, and change every occurrence of the variable and the bounds consistently.
For the standard integration-by-parts formula, take both selected functions to be continuously differentiable on each finite interval under consideration.
At each singularity or infinite endpoint, require its own finite limit. Separate divergent pieces cannot be canceled to claim ordinary convergence.

Why the method works

1. Match structure before manipulating

An inner expression beside its derivative suggests substitution; a product with a simplifying derivative suggests parts. For rational functions, divide if necessary, then factor.

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

2. Derive the product exchange

Integrate the product rule and rearrange. A useful choice simplifies the remaining integral; applying the formula mechanically may not help.

\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\,dx

3. Evaluate finite pieces, then take limits

Infinity is not an antiderivative input. Use a finite cutoff; at an interior singularity, split and test both one-sided limits separately.

\int_a^\infty f(x)\,dx=\lim_{R\to\infty}\int_a^R f(x)\,dx

Worked examples

EXAMPLE 1

Evaluate the integral using substitution and parts.

The factor x matches the derivative of the inner quadratic up to a constant. Change the endpoints as well as the differential.
$I=\int_0^1x\ln(1+x^2)\,dx,\quad u=1+x^2,\quad du=2x\,dx,\quad I=\frac12\int_1^2\ln u\,du$
Treat the logarithm as a product with 1. Differentiating the logarithm and integrating 1 leaves a constant integrand.
$\int\ln u\,du=u\ln u-\int1\,du=u\ln u-u+C$
Evaluate in the new variable. Differentiating the resulting antiderivative independently recovers the original integrand.
$I=\frac12[u\ln u-u]_1^2=\ln2-\frac12,\qquad \frac{d}{dx}\left(\frac{(1+x^2)\ln(1+x^2)-(1+x^2)}2\right)=x\ln(1+x^2)$

Conclusion: The exact value is the natural logarithm of 2 minus one half.

EXAMPLE 2

Determine whether the positive logarithmic tail has finite total accumulation.

Replace the infinite endpoint by a finite R greater than 1; only the resulting limit decides convergence.
$J=\int_1^\infty\frac{\ln x}{x^2}\,dx,\qquad J_R=\int_1^R\frac{\ln x}{x^2}\,dx$
Choose the logarithm for differentiation. Integration by parts reduces the tail to an elementary reciprocal power.
$u=\ln x,\quad dv=x^{-2}dx,\qquad J_R=\left[-\frac{\ln x}{x}\right]_1^R+\int_1^R x^{-2}\,dx=1-\frac{\ln R+1}{R}$
The logarithm grows more slowly than the linear denominator. For example, $\displaystyle L'$ Hopital's rule verifies the needed logarithmic quotient limit.
$\lim_{R\to\infty}\frac{\ln R}{R}=\lim_{R\to\infty}\frac{1/R}{1}=0,\qquad J=1$

Conclusion: The improper integral converges to 1, although the integration interval is infinite.

Pitfalls & counterexamples

Tempting claim: Symmetric cancellation always makes an improper integral converge.

Correction: Ordinary improper integration requires separate convergence on both sides of a singularity; a symmetric principal value is a different notion.

Test the claim: The reciprocal function on the interval from negative one to one has symmetric cutoff value zero, but its two one-sided integrals diverge.

Check your understanding

Is decay of a positive integrand to zero sufficient for its integral over an infinite interval to converge?

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Compare your reasoning

No.

The reciprocal function tends to zero, but its integral from one to a growing upper bound is the logarithm of that bound, which is unbounded.

Connect this idea

Improper integrals motivate comparison tests for series. Substitution becomes change of variables with a Jacobian in multiple integrals.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.