If $f(z)$ is an analytic function, then find the value of the integration $$\int_{0}^{2\pi}f(e^{it}) \cos t \, dt.$$  My Work: Taking $e^{it} = z$ the integrand becomes of the form $$i\frac{f(z)\operatorname{Re(z)}}{z}$$ on the simple closed contour $|z|=1$. But how can I proceed from here? Please help.
85755
$f$ be analytic function, could any one tell me how to find the value of $$\int_{0}^{2\pi} f(e^{it})\cos t \,\mathrm dt$$   I am not able to apply any complex analysis result here, could any one give me hint?
36956
I am attempting to find the Laurent expansion at $z=0$  for $$f(z) = \sin {\frac{1}{z}}$$ where $z$ is a complex number. I rewrote the function as $$f(z)=\frac{ e^{iz^{-1}} - e^{-iz^{-1}}}{2i}$$ $$=\frac{1}{2i} \sum_{n=0}^{\infty}[i^n-(-i)^n]\frac{z^{-n}}{n!} $$  What do I do next?
19066
Let $f$ be analytic on an open set $U$, let $a \in U$, and $f'(a)\neq 0$.  Show that  \begin{align*} \frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz \end{align*} where $C$ is a small circle centered at a.  I thought that this is very similar with Cauchy's integral formula. But, I can't apply the Cauchy's integral formula. How to apply that?   Any help is appreciated..   Thank you!
94565
There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write: $$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$  My question is: What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)  (Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)
102113
I'm wondering how to integrate the so-called integral using Residue theorem,as it has a pole of second order on the real axis(not simple) so we cannot use $\pi i Res(@ z=0)$.Would you please give me a hint?($a,b&gt;0$)
44339
I'm having trouble calculating the integral  $$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$  using residue calculus. I've previously encountered expressions of the form   $$\int_{- \infty}^\infty f(x) \sin x \,dx $$  where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$  How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work
111255
$$\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^2(x^2+16)}$$  My attempt:  We are integrating over the real axis, which is the real part of a set of complex numbers, so $$=\int_{-\infty}^{\infty} \frac{dz}{(z^2+1)^2(z^2+16)}$$  $$=\int_{-\infty}^{\infty} \frac{dz}{((z+i)(z-i))^2(z+4i)(z-4i)}$$  But I notice that all of the singularities are imaginary! Should this give me pause? What should I do now?
67111
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$      $f(x) = \begin{cases} \frac {1}{n},  &amp; \frac{1}{n+1}&lt;x\le\frac{1}{n}\\ 0, &amp; x=0 \end{cases}$   My attempt  $f$ is bounded and monotonically increasing on $[0,1]$. Also, $f$ has infinite discontinuities but only one limit point. Therefore $f$ is Riemann integrable. Now, to calculate the integration  $\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...$  $=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}$  How do I proceed from here? How do I calculate these summations? I know $\sum \frac{1}{n}$ is $\log 2$, but not the other two summations.
103339
Show $\int_C \frac{dz}{(z-z_0)^n} = 2\pi i$ when $n = 1$ and 0 when $n \gt 1$   I'm not entirely sure how to approach this one.  Any help would be much appreciated.
24681
There is a holomorphic function $f$ from the unit disc to itself s.t $f(0)=f(i/2)=0$. How can you prove the estimates $|f'(0)|\leq 1/2$ and $|f(-i/2)|\leq 2/5$?   Applying Schwarz's lemma we can get $|f(z)|\leq |z|$ but this gives an estimate too large. Cauchy's integral formula seems of no use since we don't know anything about other values of the function.
71564
Let $f$ be analytic in $|z|\ge1$ and  $\lim_{z\to\infty}f(z)=0$. Then is $f(z)=\frac1{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{z-\zeta}d\zeta$,$\forall |z|\ge2$, where $\Gamma$ is the circle of radius $2$ with centre at the origin.   I think yes, by the Cauchy integral formula. But, how do we ensure that the non analyticity of $f$ in $|z|&lt;1$ dosent affect the formula? Thanks beforehand.
101590
I need to evaluate $\int_{\gamma} \frac{ |dz| }{z} $ where $\gamma $ is the unit circle.    Attempt: Let $\gamma(t) = e^{it} $ where $t \in [0,2 \pi]$. How can I get rid of the absolute values in the differential?
106227
Suppose that  $f(z)$ is analytic and $f(z)\neq0$  on the disc $|z|&lt;1$ .  Show that for $0&lt;r&lt;1$,  $$ \exp\left[\frac{1}{2\pi}\int_{0}^{2\pi}{\log}|f(re^{i\theta})|\,d\theta\right] = |f(0)| $$  Any other theorems needed to solve this problem? I don't have any idea just with "Gauss's MVT"  Thank you in advanced.
105432
Is there a way I can derive the value of the integral  $ \int_0^1 \ln(x)\ln(1-x)dx$  using the fact that  $\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  ? (the actual value of the integral is $2-\frac{\pi^2}{6}$)  Thanks in advance
4852
I need to prove $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$ for $0&lt;Re(z)&lt;1$. I tried $\cos t=\frac{e^{it}+e^{-it}}{2}$ and to integrate it along the contour from $\epsilon $ to R, then to $Ri$ via a quarter of a circle, then downward to $\epsilon i$, finally back to $\epsilon$ via a small quarter of a circle. However, I find that $e^{-it}t^{z-1} $ does not converge when $R\to \infty$, can anyone give a correction of my integration or offer another method to evaluate the integral?(would be better if using complex contours) thanks in advance.
56658
Suppose that $f$ is analytic on a close curve γ. Prove or disprove   $$\int_\gamma \overline{f(z)}f'(z)dz$$  is purely imagine.  I know that $f$ is analytic on a close curve, then   $$\int_\gamma f(z) dz=0$$  I tried an example with $\gamma =e^{it}$ with $0\leq t\leq 2\pi$, I often get the real part of the integral equal zero.  I tried let $f=u+iv$ so $f'=u_x+iv_x$. Since $f=u+iv$, $\overline{f}=u-iv$  $$\overline{f(z)}f'(z)=uu_x+vv_x+i(uv_x-vu_x)$$  But this doesn't get me anywhere. Any help would be greatly appreciated.
11212
Consider the function,   $$f(z) = \frac{\sin(\frac{1}{z})}{z^2+a^2} \text{ where } a&gt;0$$  I know that this has two simple poles at $z = \pm ia$ and an essential singularity at $z = 0$ because of $\sin(\frac{1}{z})$ in the numerator. For $\operatorname{Res}(f,ia)$ and $\operatorname{Res}(f,-ia)$, it is relatively easy to find the these values using   $$a_{-1} = \lim_{z\rightarrow z_0} ((z-z_0)f(z))$$  However, I am finding it much more difficult to determine the residue at an essential singularity.  From what I have read, it seems that the easiest way to determine the residue is to explicitly find the Laurent Expansion and then get it from there. Would this be the case for this function as well?