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Oddness gives zero cosine coefficients. Integration by parts gives the sine Fourier coefficientsThe Fourier series of the piecewise smooth periodic extension converges to its value at interior continuity points. Termwise integration from 0 to $x$ is valid: on the interval between these points, which stays inside $(-\pi ,\pi )$, the partial sums of $\sum (-1{)}^{n+1}\sin (nt)$ are uniformly bounded by the geometric-sum formula, since $|1+e^{it}|$ is bounded away from zero. The Dirichlet test therefore gives uniform convergence there. Integrating givesThe integrated series is absolutely and uniformly convergent. ConsequentlyIndependently, the constant Fourier coefficient is $a_0={\pi }^{-1}{\int }_{-\pi }^{\pi }{x}^{2}dx=2{\pi }^2/3$. Comparison yieldsThe original sine series is not uniformly convergent on the whole open interval; that stronger assertion is unnecessary for the justified integration.

The series for $S$ converges uniformly by the M-test. Therefore termwise integration is valid and every term integrates to $\sin (n^3\pi )/n^7=0$, so the integral is zero.

For $|t|<1$, the geometric series gives $(1+t{)}^{-1}={\sum }_{n\ge 0}(-t{)}^n$. Its uniform convergence on every closed interval strictly inside $(-1,1)$ permits termwise integration from zero to $x$. ThereforeSubtracting cancels every even power and doubles every odd power, giving the Taylor series

The power series for $e^{y^3}$ has uniform convergence on every compact interval, so termwise integration from zero to $x$ is valid. ThusThis power series converges for every real $x$, including negative $x$ with the usual orientation of a definite integral.

The volumetric flow rate is the integral of axial velocity over the cross-section:The plane Poiseuille flow contribution isFor each correction eigenfunction,Uniform convergence permits termwise integration. Multiplying these two integrals by $A_n$ givesConsequentlyEvery summand in the subtracted quantity is positive. Thus the side walls reduce the flux below the infinite-width plane Poiseuille flow prediction. Its dimensions are volume per unit time, as required for a volumetric flow rate.

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

For real $a$, the geometric-series expansion of $1/(e^x-1)$ gives an integrable series after multiplication by $\sin (ax)$: each absolute integral is bounded by $|a|{\int }_0^{\infty }x{e}^{-nx}dx=|a|/n^2$. Termwise integration is therefore justified and gives the convergent series ${\sum }_{n\ge 1}a/(a^2+n^2)$.