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4.10.6 · D5Advanced Topics (Elite Level)

Question bank — Residue theorem — computing real integrals

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Figure — Residue theorem — computing real integrals

True or false — justify

Every line is a claim. Decide true or false, then give the one-sentence reason.

The residue theorem needs to be analytic everywhere inside the contour.
False. It is analytic inside except at the isolated poles — those exceptions are exactly what produces the term; a fully analytic function would give by Cauchy's Integral Theorem & Formula.
A residue is a property of the point, not of the contour you draw around it.
True. is fixed by the Laurent series at ; the contour only decides whether that residue gets counted, not its value.
If , the semicircle formula still works.
False. On the arc the height is and the arc length is , so the bound is — a nonzero constant that does not vanish, so the boxed formula is invalid.
For you could equally close in the lower half-plane.
False. With , decay only happens for (UHP); in the LHP explodes, so Jordan's Lemma fails there.
Closing a real integral in the UHP versus the LHP gives the same answer.
True (for a valid rational integrand) — the value is the same, but LHP uses (clockwise orientation) and the lower poles, so the arithmetic differs while the result agrees.
The residue at a removable singularity is always zero.
True. A removable singularity has no negative-power terms, so ; the point contributes nothing to any contour sum.
Every function with has an antiderivative near .
True locally. With every Laurent term integrates termwise to a single-valued function, so a local primitive exists in the punctured disk.
An essential singularity has no residue.
False. It still has a Laurent series and hence a ; the residue theorem applies just fine — you just can't use the tidy pole formulas.
requires you to check decay at infinity.
False. The substitution turns it into a finite loop on ; there is no arc going to infinity, so no decay condition — only "which poles sit inside the unit circle."

Spot the error

Each line states a plausible-looking move. Say what's wrong.

" has poles at , so I sum both residues: ."
The UHP contour only encloses ; is outside the semicircle, so it must not be summed — including it wrongly gives instead of .
"For I use and take residues."
contains , which blows up in the UHP; use instead and take the real part at the end.
"Trig sub: ."
The factor was dropped; without the integrand is wrong and you also miss the possible extra pole at .
"."
The pole is order 2, so you must multiply by and differentiate once; multiplying by only leaves a divergent limit.
"Since has , I take the residue of ."
Same trap as : use and take the imaginary part; itself grows in the UHP.
" works for any pole."
Only for a simple pole (single zero of with ); for higher-order poles the derivative formula with is required.
"The arc vanishes, so I can drop it before checking the degree gap."
You must verify the gap () or Jordan's condition first; assuming the arc dies is the single most common way the boxed formula gives a wrong number.
" is a pole, so I include its residue for ."
That root has , so it lies outside the unit circle and is not enclosed; only counts.

Why questions

Give the reason, not just the fact.

Why is only the coefficient called the residue and not any other ?
Because integrating around a small circle gives for every and for — is the only term that survives the loop integral.
Why do we close in the upper half-plane by default for rational integrands?
Convention only — either half works for rationals since decay is symmetric; we pick UHP so we can reuse the same setup when Jordan's Lemma (which requires the UHP) enters for oscillatory integrals.
Why does Jordan's Lemma let us succeed with only a degree-1 gap when the plain semicircle needs a degree-2 gap?
The factor decays exponentially on the arc, overpowering the mere from , so we no longer need itself to decay fast — see Jordan's Lemma.
Why does the trig substitution sometimes introduce a pole at ?
Because carries a factor , and bring negative powers of too — these can create a genuine singularity at the origin that must be checked.
Why must the poles be isolated for the residue theorem to apply?
A non-isolated singularity (a limit point of singularities, or a branch cut) has no single Laurent series, so no well-defined — such cases need Contour Integration & Branch Cuts instead.
Why does an odd integrand like confirm the imaginary part answer without extra work?
An odd function integrated symmetrically over gives , which matches the imaginary part when it should vanish — a free consistency check on the residue arithmetic.
Why can't we use the residue theorem directly when a pole sits on the real axis?
The contour would pass through the singularity, so the integral is not even defined in the ordinary sense; you must indent around it and use Principal Value Integrals.
Why does multiplying the residue sum by (and not just ) give the closed-contour value?
The full circle sweeps at ; a half-circle would give , which is exactly why indented semicircular detours around real poles contribute half-residues.

Edge cases

Boundary and degenerate scenarios — state precisely what happens.

What is if has no poles in the UHP and ?
With the decay hypothesis the arc vanishes, so the empty residue sum gives exactly ; without the degree gap the arc survives and this conclusion fails — the hypothesis is essential, not optional.
What happens to the formula if a pole lies exactly on the semicircular arc as ?
For a fixed finite set of poles this never occurs for large (poles are bounded), so it's a non-issue — you just take larger than the biggest .
If two poles coincide (a repeated root of ), does the theorem break?
No — they merge into one higher-order pole, and you use the order- derivative formula; the theorem is unchanged, only the residue computation gets harder.
What is the residue at a simple pole of for the trig-substitution case if it lands on ?
That means the original real integral had a where -type blow-up; the integrand is singular on the path, so you'd need a principal-value treatment — the plain formula fails.
For , what happens as ?
It smoothly tends to , matching ; the exponential decay from the residue becomes .
What if in ?
Then decays in the lower half-plane, so you must close downward (clockwise, factor ) and use LHP poles — closing up would make the arc blow up.
If the degree gap is exactly 2 but the leading coefficients make decay only like , does the arc still die?
Yes — , so degree-2 is the sharp threshold: it works, degree-1 does not.
Recall One-line self-test

Cover the answers and race through the "Spot the error" section; if you can name the fix for each in under ten seconds, you've internalised the traps.

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