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For a counterclockwise simple contour containing $z_0$, and $f$ analytic on a neighbourhood of its interior and boundary, the Cauchy integral formula isFor the required rational integral, use partial fraction decomposition:Both points $1,-1$ lie inside $|z|=2$. Applying the formula with the constant analytic function one gives $2\pi i$ for each of the two terms. They cancel, soReversing the circle's orientation would leave this zero result unchanged.

Partial fraction decomposition give $h(z)=1/(z-i)+1/(z+i)$. Let $w=z-1$. Each fraction has a geometric expansion, soEquivalently, its coefficient is $(-1{)}^j{2}^{(1-j)/2}\cos ((j+1)\pi /4)$. The first terms areThe linear and fifth-order coefficients vanish. Both geometric series converge for $|w|<\sqrt{2}$, and $h$ has genuine simple poles at $z=\pm i$, each distance $\sqrt{2}$ from the centre. Thus the Taylor series has

Set $w=z-1$. Partial fraction decomposition givesExpand both terms as geometric series:The two bracketed terms are complex conjugates. Using $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$ produces the required real coefficients:The initial terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$. The nearest singularities are the simple poles $z=\pm i$, both at distance $\sqrt{2}$ from the center. Hence the Taylor series has

Put $a=1/L\in (0,1)$. The Fourier transform of a derivative turns the differential operator into $k^4+5k^2+4=(k^2+1)(k^2+4)$, soUse the partial fraction decompositionThe inverse-transform pair ${\mathcal{F}}^{-1}[(k^2+b^2{)}^{-1}]=e^{-b|x|}/(2b)$ is the one-dimensional modified Helmholtz Green function. HenceIn terms of the original length scale,This is the convolution of $h$ with the fourth-order Green kernel with two decay scales, here $K(x)=e^{-|x|}/6-e^{-2|x|}/12$. It decays at both infinities. Although the individual exponential pieces have cusps, their derivative jumps through order three cancel, so the result has no unintended Dirac delta distribution at zero. The decaying solution is unique: the homogeneous solutions $e^{\pm x},e^{\pm 2x}$ have no nonzero globally decaying combination.

Take the causal solution, extending the input and output by zero before the device starts at $t=0$. The two zero initial conditions ensure that this extension adds no impulsive terms to the first two distributional derivatives. The Fourier transform of a derivative is multiplication by $i\omega$, soFactoring the quadratic and taking partial fraction decomposition yieldsThe inverse Fourier transform follows directly by integrating decaying exponentials:where $H$ is the Heaviside step function. This damped-oscillator impulse response has $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, as required for a unit impulse.

Put $w=z-1$ and $a=1-i$. Partial fraction decomposition givesBoth terms admit a geometric series at $w=0$. The local Laurent series is actually a Taylor series, since the center is regular:The nearest pole is $z=i$, a distance $|1-i|=\sqrt{2}$ from the center; the other pole is at distance $4$. Thus the requested local radius of convergence is $\boxed{R=\sqrt{2}}$.

For completeness, specifying a Laurent series without a region can also mean either of the other maximal annuli. Expanding the first fraction in inverse powers and the second in positive powers givesExpanding both fractions in inverse powers instead givesThese are different Laurent series on different annuli, rather than extensions of the local Taylor series beyond its radius of convergence.

Start with the partial fraction decompositionFor $1<|z|<2$, expand the first fraction in negative powers and the second in nonnegative powers using the geometric series:Thus the Laurent expansion on this annulus isFor $|z|>2$, both fractions expand in negative powers:Combining the two geometric series givesThe $z^{-1}$ coefficient is zero, consistently with the original rational function being $O(z^{-2})$ at infinity. Each displayed series converges absolutely on its stated annulus and uniformly on every compact subset of that annulus.

For $c_1=2$, $c_2=4$, $c_3=5$, the friction memory kernel and its Laplace transform areUsing the generalized fluctuation-dissipation relation with equilibrium velocity variance $5$, the velocity autocorrelation function has Laplace transformAll three poles are negative, so the stationary relaxation is stable. A partial fraction decomposition givesTaking the inverse Laplace transform givesThe negative middle coefficient is compatible with an autocorrelation function; individual terms in a partial fraction decomposition need not themselves be separate physical modes with positive weights. Several useful checks follow directly:The Volterra integral equation for $\chi$ predicts exactly ${\chi }^{\prime }(0)=0$ and ${\chi }^{\prime \prime }(0)=-\kappa (0)\chi (0)=-10$. Its integral is also $8-10/9+1/18=125/18=5/\tilde{\kappa }(0)$, agreeing with the Green-Kubo formula for diffusion.

The partial fraction decompositionmakes the sum a telescoping sum. All interior terms cancel:

For a partial fraction decomposition, writeMultiplying by $(x-1{)}^2(x+1)$ gives $4=A(x^2-1)+B(x+1)+C(x-1{)}^2$. Setting $x=1,-1$ gives $B=2,C=1$, and comparing the coefficient of $x^2$ gives $A=-1$. ThusIntegrating each term gives the antiderivativeThe constant of integration can be chosen independently on each interval separated by $-1$ and $1$.

For $x\ne \pm 1$, write the linear ordinary differential equation asAn integrating factor on each interval avoiding $\pm 1$ is $\mu (x)=1/(1-x^2)$; multiplication by a nonzero constant, including a sign, is immaterial. ThereforeThe partial fraction decomposition isIntegrating gives the general solutionThe constant is independent on each connected interval of definition. At the singular points the undivided differential equation forces $y(1)=0$ and $y(-1)=1$. The displayed family tends to these values, but its derivative has a logarithmic divergence there, so no member extends as a differentiable function through either singular point.

Use partial fraction decomposition:Integrating each term gives the antiderivativeThis holds on any interval avoiding $-2,0,2$; the constant may differ between these intervals. Differentiation of the logarithm recovers $1/[4(x^2-4)]$, providing a direct check.

Use partial fraction decomposition:Integrating givesThis real antiderivative is valid on each interval avoiding $x=\pm 2$, with its own integration constant. In particular, $I(1)-I(0)=\frac{1}{4}\log 3$.

For $n=1$, both sides are $1/3$. Assuming the formula through $n$, adding the next term givesThe last expression is the required formula at $n+1$, completing mathematical induction. The partial fraction decompositionprovides a second explanation: this is a telescoping sum.

Let $P$ be a monic polynomial of positive degree $m\ge 1$ with distinct roots of a polynomial $r_j$. The partial fraction decompositionfollows by matching the simple pole residues; the difference is entire and vanishes at infinity. Comparing its expansion there gives ${\sum }_j{r}_j^{\ell }/P^{\prime }(r_j)=0$ for $0\le \ell <m-1$ and ${\sum }_j{r}_j^{m-1}/P^{\prime }(r_j)=1$. These identities give the continuity and derivative jumps needed for constant-coefficient Green functions.

A series adds the terms of a sequence in their specified order. Its infinite-sum notation is defined through finite partial sums:when the displayed limit of a sequence exists as a finite number. Otherwise the series is not a convergent series in the ordinary sense. Writing an infinite list of additions does not by itself assign it a finite sum. Cesàro summation is a different prescription, with its own definition.

Two positive series make the distinction concrete. For $a_n=1/n$, the individual terms tend to zero but the harmonic series has unbounded partial sums. For $a_n=1/[n(n+1)]$, partial fraction decomposition reveals a telescoping series whose partial sums tend to one. The useful question is how much all the remaining terms can contribute together, rather than only how small the next term looks.

A telescoping series has partial sums in which internal contributions cancel. The basic pattern is $a_n=b_n-b_{n+1}$, for which finite summation givesOnly the two boundary terms survive. If $b_n$ tends to a finite limit of a sequence $L$, the series consequently sums to $b_1-L$. The cancellation is exact for each finite $N$; the infinite result then follows by a limit of a sequence.

For example, partial fraction decomposition givesThe first three terms are $1/2,1/6,1/12$, with partial sum $3/4$. The first five have partial sum $5/6$. Taking the limit of a sequence gives the concise answer

The appeal is that a long calculation becomes a boundary calculation. It works whenever the summand is a successive difference, not only for rational functions. For instance, $a_n=\log ((n+1)/n)=\log (n+1)-\log n$ gives partial sum $\log (N+1)$, so its series diverges despite the cancellation. A telescoping series therefore need not be a convergent series: one must still examine the surviving boundary term.