Articles of complex integration

Prove the following equality: $\int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi$

This question already has an answer here: Evaluate $\int_0^{\pi} e^{a\cos(t)}\cos(a\sin t)dt$ 4 answers

Difficulty evaluating complex integral

The integral along the path $\gamma(t)=e^{2ti},\;t\in[0,2\pi]$ is $\begin{equation*} \int_{\gamma}\frac{1}{z^{2}-1}dz \end{equation*}$. I approached this like a real integral in the hopes things would work out, first by performing a u sub $\begin{equation*} u=e^{2ti}\Rightarrow du=2e^{2ti}dt\Rightarrow dt=\frac{du}{2e^{2ti}} \end{equation*}$ Which brought me to $\begin{align*} &\int_{t=0}^{t=2\pi}\frac{i}{u^{2}-1}du= i\int_{t=0}^{t=2\pi}\frac{1}{(u+1)(u-1)}du\\ &=i\frac{1}{2}\int_{t=0}^{t=2\pi}\frac{1}{(u+1)}du- i\frac{1}{2}\int_{t=0}^{t=2\pi}\frac{1}{(u-1)}du\end{align*}$ In the reals this would obviously $\frac{i}{2}\log(e^{2ti}-1)-\frac{i}{2}\log(e^{2ti}+1)|_{t=0}^{2\pi}$ but i am pretty […]

Cauchy's Integral parametric conjugate

By considering the conjugate of its parametric form, evaluate $$\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{\overline{f(z)}}{z-a}dz$$ when $|a|<1$ and $|a|>1$, where $f$ is holomorphic in in the disk $(0;R), R>1$. Typically when doing these kinds of integration and parametrization, $|z|=n$ is given, but it’s different in this case (or is it not?). Can someone help me out?

evaluate $\oint_C \frac{1}{z-i} dz$ where C is the circle $\left\vert 2 \right\vert$

I’ve tried several times to solve this integral, but I am unable to solve it, every time I get a similar wrong answer. The question is “evaluate $$\oint_C \frac{1}{z-i} dz$$ where C is the circle $\left\vert 2 \right\vert$” Here is my (wrong) solution: This function is not analytic in $z=i$. This point is part of […]

Compute $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$

I”m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. Here it goes: We want to evaluate $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$. Here’s my work: Let $z=e^{i\theta}$ so that $d\theta=dz/iz$ and $\cos \theta = \frac12 (z+z^{-1})$. We have $$\begin{align} I&=\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta \quad (1)\\\\ &=\oint_C \frac{1}{(2+\frac{z+z^{-1}}{2})^2}\frac{dz}{iz} \quad (2)\\\\ &=\frac1i\oint_C \frac{4}{(4+z+z^{-1})^2}\frac{dz}{z} \quad (3) \\\\ &=\frac4i\oint_C \frac{1}{(z^2+4z+1)^2}\,dz \quad (4) […]

Integrate complex conjugate

I’m doing some exercises on complex integration, and stumbled over this: $$\int_\gamma \bar{z} dz,$$ Where $\gamma$ is any closed $C^1$ curve which is the boundary of a bounded, connected, open $U \subset \mathbb{C}$, and $\bar{z}$ denotes the complex conjugate of $z$. One should express the result in terms of the quantity $$\int_\gamma x dy ,$$ […]

Evaluate $\int_{|z|=r}^{}\frac{\log\ z}{z^2+1}dz$ where $r>0$

I tried to evaluate $$\int_{|z|=r}^{}\frac{\log\ z}{z^2+1}dz$$ when $r>0$. Clearly $\log z$ is not continuous at $z=-r$. So is this integral meaningful then? Since the function is continuous and bounded on the given path except only at this point, I am thinking it should be possible to actually evaluate this. Hope someone could help me out […]

How to integrate this improper integral.

I am trying to integrate this: $$\int_0^\infty\frac{ 1}{x^2}$$ I was trying to convert it into a complex integral. But did not know how to proceed. My original question is : $$\int_0^\infty\frac{1-cosx}{x^2}$$. Can someone give me a hint on how to proceed. Thanks in advance.

question related with cauch'ys inequality

We have a statement that if f(z) is analytic in a domain D and $C = \{z : |z-a| = R\}$ is contained in D. Then, $|f^n(a)|\leq \frac{n!M_R}{R^n}$ where $M_R = \max|f(z)|$ on C. What if the given D is not a circle? Then how to find the bound? I had problem in solving the […]

Find the principal value of $\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$

How to find the Cauchy principal value of the following integral $$\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$$ How to start this problem?