Real integrals and contour estimates

Try

The integral of 1/(x²+1) over the real line is π. A complex contour explains the same value through the single upper-half-plane pole at i. Close the real segment from −R to R with a counterclockwise upper semicircle, calculate the residue, and then show the added arc contributes nothing in the limit.

The last step is essential. A residue calculation gives the integral around the whole closed contour, not automatically the integral along the real segment.

Predict

Predict the arc bound for 1/(z²+1). On |z|=R>1, the denominator modulus is at least R²−1 and the arc length is πR. Their ratio tends to zero.

Now replace the integrand by one. The same arc has integral that does not vanish. A large contour alone does not make an auxiliary contribution negligible; decay of the integrand must outweigh the growing length.

Experiment

Track the three parts of the rational calculation.

Quantity	Formula
Residue at i	1/(2i)
Closed contour integral	π
Upper arc modulus bound	πR/(R²−1)

Once the bound tends to zero, the real segment integral tends to π. Absolute integrability of 1/(1+x²) makes this the ordinary improper integral, not merely a symmetric principal value.

Understand

For a>0, the function 1/(z²+a²) has upper pole ia with residue 1/(2ia). The residue theorem gives π/a for the closed contour. On the upper semicircle, |z²+a²|≥R²−a², so the arc integral is bounded by πR/(R²−a²), tending to zero. Therefore the real-line integral is π/a.

Oscillatory factors require checking their behavior in the chosen half-plane. For b>0, |exp(ibz)|=exp(−b Im z)≤1 in the upper half-plane. Thus the same elementary arc bound works for exp(ibz)/(z²+a²), without needing a stronger named arc lemma. The residue at ia is exp(−ab)/(2ia), yielding πexp(−ab)/a. Taking real parts gives the cosine transform; the sine contribution vanishes by oddness on the real line.

For rational functions with faster decay, estimate the denominator by its leading term uniformly on a large circle and include every pole in the chosen half-plane. If a pole lies on the real axis, the usual contour is invalid. One must specify an indentation and calculate its contribution with its actual orientation.

An ordinary improper integral requires the separate one-sided limits to exist. A Cauchy principal value instead uses a specified symmetric limiting procedure. For 1/x on [−1,1], symmetric truncations cancel to zero, while each one-sided improper integral diverges. The two notions cannot be interchanged by a contour picture.

Apply

Evaluate the upper poles of 1/(z⁴+1): they are exp(iπ/4) and exp(3iπ/4). Their residues are 1/(4z³). Sum them before multiplying by 2πi, and verify the result π/√2 is real and positive, as the original real integrand requires.

Diagnose a suspicious answer with a nonzero imaginary part. It may indicate a missing pole, an orientation error, or incorrect conjugate arithmetic. A qualitative property of the original integral is a valuable independent check.

Continue

Residues also count zeros through the logarithmic derivative, which is the next topic. Optional depth: derive a small-indentation contribution directly with z=a+εexp(it), retaining the direction and parameter interval. Do not use an automatic “half residue” rule without specifying which side of the pole and which traversal the contour uses.