# Tagged Questions

Questions on the evaluation of residues, on the evaluation of integrals using the method of residues or in the method's theory.

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### The meaning of the Imaginary value of the Residue while Evaluating a Real Improper Integral

When evaluating the improper integral $$\int_{0}^{\infty}\frac{x^{3}\sin\left(2x\right)}{\left(x^{2}+1\right)^{2}}\,dx$$ (which is an even function, so half of the $(-\infty,\infty)$ integral), I used ...
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### Calculating the residue of a complex funciton with ln(z) at z=0

How can I calculate this residue: $$Res\left(\frac {z\ln(z)}{(z^2 +1)^3} , 0\right)$$ if it's possible at all. I know $0$ is a branch point for $\ln(z)$ and therefore isn't a pole, but when i plug ...
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### Inverse Laplacetransform of rational function with multiple pole

I have to calculate the inverse Laplacetransorm of this function using Residue calculus $$\frac{s^4 + 6s^3 - 10s^2 + 1}{s^5}$$ but I can't find any Residue rule that would solve this. Can you show ...
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### Why is the (-1)-th coefficient of $f^n f'$ equal to 0, without dividing by $n+1$?

Let $R$ be a commutative ring, and $n$ be a nonnegative integer. Let $f\in R\left[t,t^{-1}\right]$ be a Laurent polynomial in one variable $t$ over $R$ (this means a formal $R$-linear combination of ...
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### Multivariate Residue Theorem

Let $G(s,t)$ be a complex valued function in two variables that converges absolutely for $Re(s), Re(t)>1$. Suppose we can analytically continue $G$ in such a way that $$G(s,t) = f(s)g(t)H(s,t)$$ ...
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### Complex Analysis - Calculating Residues

I am told to calculate the residue of $\frac{ e^{-z} }{ (z-1)^{2} }$ at $z = 1$. The answer is supposed to be $\frac{ 1 }{ e }$. Can someone give me a hint on how to approach this?
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### Compute the following integral, where $C$ is the circle $|z|=3$

Evaluate:$$\int_{C} (1 + z + z^2)(e^\frac{1}{z}+e^\frac{1}{z-1}+e^\frac{1}{z-2}) dz$$ where $C$ is a circle $|z|=3$ and $z \ \epsilon \ \mathbb{C}$ The function that is being integrated has ...
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### Why does this example of global residue theorem not work?

This question is related to and inspired by a previous question What is the residue obtained from this integral? , but note that the appearing functions are slightly different. Consider the following ...
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### Finding a generalization for $\int_{0}^{\infty}e^{- 3\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(3\pi x)}dx$

$\;\;\;\;$I was reading the introduction of Paul J. Nain's book "Dr. Euler's fabulous formula" where he talks about the sense of beauty in mathematics and quotes the G.N.Watson as saying that a ...
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### Complex analysis, poles and singularities and boundedness

So I am on the following problem: Prove that an isolated singularity of $f(z)$ is removable as soon as either $\text{Re}f(z)$ or $\text{Im}f(z)$ is bounded above or below. The hint is to use a ...
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### $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}$ using residues

I'm trying to find $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx$ where $\alpha>0$ is real. My approach was to take an integral along the real line from $1/R$ to $R$, around the circle ...
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### Evalulate $\int_{-\infty}^{\infty}\frac{1}{(1+x^{2n})^2}dx$ by using residue theorem

I know the answer of the integral $$\int_{-\infty}^{\infty}\frac{1}{1+x^{2n}}dx=\frac{\pi}{n\sin\left(\frac{\pi}{2n}\right)}$$where $n\in\mathbb{N}$. But how to evalulate ...
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### Is there a simpler way to compute the residue of a function at a pole of order 3?

The function $$\frac {1}{z^2(e^{i2\pi z}-1)}$$ has a triple pole at z = 0. To compute the residue of f at z = 0, I can compute the Laurent expansion of f about z = 0, and then read off the ...
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### Computing residues of $\cot(\pi z)/z(z+1)$ with symmetries

I would like to know if there is a quick way of computing the residues of $$f(z) = \frac{\cot \pi z}{z(z+1)}$$at the points $z = 0$ and $z = -1$. They are double poles. Expanding this in Laurent ...
I am having a bit of trouble evaluating $$\sum_{k=1}^3{ \rm Res}\left(\frac{\log(z)}{z^3+8};z_k\right)$$ where $z_1=2e^{i\pi}$, $z_2=2e^{i\pi/3}$ and $z_3=2e^{i5\pi/3}$. I know that each $z_k$ is a ...