If $C$ is a contour in the complex plane, is there a geometric interpretation of the contour integral $\int_Cf(z)dz$? What does the value of this integral mean/tell us about anything? Why does the sign of the integral change when the contour is traversed in the opposite direction?
122438
Im trying to get my head around complex integration/complex line integrals. Real integration can be thought of as the area under a curve or the opposite of differentiation. Thinking of it geometrically as the area under a curve or the volume under a surface in 3 dimension is very intuitive.   So is there a geometric way of thinking about complex integration? Or should I just be viewing it as process that reverses differentiation? Or has integration other meanings in complex analysis? Here is an example, could someone explain this to me - Here's the definition of the integral along a curve gamma in C, parameterized by $w:[a, b] -&gt;C$ \begin{equation} \int_\gamma f(z)dz = \int_a^b f(w(t)).w{(t)}'dt \end{equation}   So I have -  $\gamma$ is the unit circle with anti-clockwise orientation parameterized by  $w:[0, 2\pi]\to C$  $w(t) = e ^{it} = Cos(t) + iSin(t)$  So if use the definition of the integral,   $\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}&#39;dt$,  and work this out it comes to  $\int^{2\pi}_0 i dt = 2{\pi}i$  So what does this $2{\pi}i$ represent? Does it mean anything geometrically, like if a regular integral works out to be 10 that means the area under the curve between 2 points is 10...
5470
I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.  Here is a version of a rational log integral rarely encountered.  $\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$  I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.  My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?.  Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?.
88402
I am trying to evaluate $$\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx $$ Would a contour work? I have tried using a contour but had no success. Thanks.  Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways.
110254
Let $f(z)$ be analytic in a closed set $A$ and bounded by contour $C$. Let $z_1,z_2,...,z_n$ be different arbitrary point in the interior of $C$ and let $w_n(z)=(z-z_1)(z-z_2)...(z-z_n) $. Prove the integral $$P(z)=\frac{1}{2\pi i} \int_C \frac{f(x)}{w_n(x)} \frac{w_n(x)-w_n(z)}{x-z} dx$$ is a polynomial of degree $(n-1)$ that is the same as $f(z)$ at $z_1,z_2,...,z_n$ points.  Note: P(z) is the Lagrange interpolating polynomial.  I think I have an idea for the proof, but I don't know how to write it formally.  Idea I think this formula $$f^n(z_0)=\frac{n!}{2\pi i} \int_J \frac{f(x)}{(x-z_0)^{n+1}} dx $$ will be useful because practically is telling us that P$(z)$ must be 'equal to' $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$.  Now the rest is trying to proof that $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$ must be something very similar to the Lagrange polynomial.  Is this a good idea? Could be demonstrated in another way? Thanks for any suggestions/hint/tip/help.
42049
let $\gamma$ be a closed continuosly differentiable path in the upper half plane  not passing through $i$. What are the possible values of the integral $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{z^2+1}dz$$ well the integral can be broken like  $$\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{(z+i)(z-i)}dz=$$  $$\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-i}dz-\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z+i}dz=$$ by Cauchy Integral Formuale $$f(i)-f(-i)$$,so the second integral is $0$ as it is analytic in the upper half plane, but the first integral iss just $n(\gamma,i)$, which iss the windding number of $\gamma$ around $i$ what more I can say? Thank you.
26163
I'm currently studying integrals in calculus. In teaching improper integrals over infinite intervals, I come across this example in my lecture notes,  $$\int_{-\infty}^{\infty} x \ dx$$  Naturally, the integration is split into two halves, such as $\int_{-\infty}^{0}x\ dx=-\infty$ and $\int_{0}^{\infty}x\ dx=\infty$. The lecture notes conclude that since both these improper integrals diverge, then so does the above improper integral, i.e.  $$\Rightarrow\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}$$    Intuitively, however, I would expect that the first improper integral, $\int_{-\infty}^{\infty}x\ dx$ should evaluate to $0$, given that,   An intuitive meaning of the definite integral is the area under the curve The curve of $y=x$ gives a negative area for $(-\infty,0)$ and positive area for $(0, \infty)$ The curve of $y=x$ is symmetrical about $y=0$   So, my questions are   Why is this not the case, and Is $\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}$ even what I think it means, that $\int_{-\infty}^{\infty}x\ dx=\infty$?
95905
I want to evaluate the integral $$\int_{\left | z \right |=3} \frac{e^{z}-1}{z(z-1)(z-i)}\,dz$$    by using the following  theorem:      If a function $f$ is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed curve $C$, then    $$\int_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).$$   I want to calculate this integral  by calculating the residue at 0 of $\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}}$ So I did long division but the coefficient of 1/z is 0.  I know the method of calculating the residues at each singularities 0, 1, i, respectively. This method gives me the answer $2\pi i(0+\frac{e-1}{1-i}+\frac{e^{i}-1}{i(i-1)})$ which is not 0. Why answers are different?? Why does not hold long division method? Any help please.
26610
I'm trying to compute $\displaystyle \int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}dx$, $(0&lt;a&lt;1)$ Let $f$ denote the integrand.  I'm using the rectangular contour given by the following curves: $c_1: z(t) = R+it, t \in [0, 2\pi]$ $c_2: z(t) = -t+2\pi i, t \in [-R, R]$ $c_3: z(t) = -R + i (2\pi - t), t \in [0, 2\pi]$ $c_4: z(t) = t, t \in [-R, R]$    There is one singularity within the contour, at $z = \pi i$. Expanding out the denominator as a power series shows that it's a simple pole, and allows us to evaluate the residue as $\displaystyle \lim_{z \rightarrow \pi i} f(z)(z-\pi i) = - e^{a \pi i}$    This is computed by expanding $1+e^z$ as a Taylor series around $\pi i$. The first coefficient will be 0, and the second will be $-1$. The rest will have orders of $(z - \pi i)$ greater than 1, and will thus vanish when we take the limit.  So the integral over the entire contour is $- 2\pi i e^{a \pi i}$    An easy enough estimate on the $c_1$ shows that the integral vanishes as $R \rightarrow \infty$. With a variable change, c_3 is the same as c_1 and also vanishes. $c_4$ becomes the integral we want when we take a limit. $c_2$ becomes $c_4$ with a constant:    \begin{align*} \int_{c_2} f(z)dz &amp;= \int_{-R}^{R} \frac{e^{-at}e^{a 2\pi i}}{1+e^{-t}e^{2\pi i}}dt \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{-at}}{1+e^{-t}}dt \\ &amp;=e^{a 2 \pi i}\int_{R}^{-R} - \frac{e^{au}}{1+e^{u}}du \ \ \ (u = -t, du = -dt)   \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{au}}{1+e^{u}}du \\ &amp;= e^{a 2 \pi i} I(R) \end{align*}    Where $I(R)$ is the line integral over $c_4$. Putting it all together and taking the limit gives us $\displaystyle \lim_{R \rightarrow \infty} I(R) = \frac{- 2\pi i e^{a \pi i}}{(1 + e^{a 2 \pi i}) }$    But this can't be the value of the integral, because it's a real-valued function integrated over $R$. I can't figure out where I'm going wrong. Note that I've avoided posting all the details of my solution since this is from a current problem set for a class on complex analysis.
33619
Well, after spending hours on this problem, I'm still stuck, so I thought I'd turn to you guys. The problem statement is as follows.  Let $f$ be a complex-valued function that is $C^1$ in the disk $|z - z_0| &lt; R$ for some $R&gt;0$, and let $C_r$ be the circle $|z - z_0| = r$. Show that $$\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0).$$  I tried parametrizing the circle in the usual way and differentiating to see if I get something useful, but no luck. I just can't get the LHS to give me something to do with $\frac{\partial f}{\partial \bar{z}}$, so that I could then put the RHS under the integral and show that the difference between the two can be less than arbitrary $\epsilon &gt; 0$ by the continuity of partial derivatives.  Thanks in advance for your help!
73606
$\int_{-2}^0\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} dydx$  y describes the circle of radius 2, and is positive and negative, so doesnt limit any quadrants in our integral, however x can only be negative, which means the left half of our circle is what we are interested in.  This means, we need the integral:  $\int_{\pi/2}^{3\pi/2}\int_{0}^{2} rdrd\theta$  Giving us a value of $2\pi$.  Is this a correct assessment? I wasnt sure if making  $0$ and $2$ the limits of $r$ was entirely correct, since it describes the entire circle of radius $2$, and not just one half. Then again, a circle of radius $2$ does have an area of $4\pi$ which makes me feel I am along the right lines.   Any clarification would be fantastic. Thank you!
40391
I have a complex integral $$\int_{|z|=r}x \, dz$$ for the positive portion of the circle.  I know that this integral seems easy enough but I am having trouble with it, and I'm fairly certain my answer is incorrect.  I wanted to use a parameter such that $x=r\cos\theta$, so we then have $dz=dx+i\,dy=dx$ since there is no reason to have have the $i\,dy$ since the imaginary part is equal to $0$.  So now I have the double integral  $$\int^r_0\int^\pi_0 r \cos\theta \, dx$$  which equals $0$ since integration the $\cos\theta$ gives $\sin\theta$ and evaluated from $0$ to $\pi$ makes the integral evaluate to $0$.  Am i misinterpreting part of the question?  Way off base?  The book suggests to also solve the problem by realizing $x=\frac{1}{2}\left(z+\frac{r^2}{z}\right)$ along the circle.  For this portion of the problem I assume that I would integrate the above by $z$ as if it were a normal calculus problem, but I have problems with what to put as upper and lower limits.  Thanks in advance!
119970
$\displaystyle\int_0^0\frac 1x\:dx$ This integral is defined ? If it define, what is the value ?  Please,step by step
115487
Is the 1 dimensional projective line homeomorphic to the circle?  If so, the circle is homeomorphic to itself with antipodal points identified  (very unintuitive).  I am missing something?
52199
If you go to 's YouTube channel and scroll through some of his videos you see him solving the following integral:  $$\int x^{dx}-1$$  he explains that this is a Product integral. My questions are the following:   1 - What is the geometric meaning of a product integral?  2 - does it make sense to have:  $$\int f(x,dx)$$  and if $f(x,dx) = g(x)dx$ then it's just a regular integrals and if $f(x,dx) = g(x)^{dx}$ it's just a product integral?  I'll leave the link to the video .
1143
I have been trying to analyse the meaning of higher order derivatives and their geometrical significance.  Given a function $f(x)$ what are the unique geometric interpretation of its higher orders?   $f'(x)$ - slope (rate of change of x w.r.t y) $f''(x)$ - convexity / concavity of $f(x)$ based on $sign(f''(x))$ $f'''(x)$ - increasing / decreasing slope of $f'(x)$ $f''''(x)$ - ?? ... $f^n(x)$ - ??   I apologize if this question is rudimentary.
12572
Let $C$ denote the unit circle  centered at the origin in Complex Plane  What is the value of   $$ \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,$$ where the integral is taken anti-clockwise along $C$?   0 1 2 3   What I have answered is 0 because it seems like $f(z)$ is analytic at 0 hence by Cauchy's Theorem.
123196
Let $f$ be holomorphic in an open $\Omega \subset \mathbb{C}$ and $\gamma$ a closed curve in $\text{int}(\Omega)$, along which $f$ is never zero.   Are these hypotheses enough to claim $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is an integer? If not, what are the necessary and sufficient conditions for that?  I can prove that's an integer for some particular cases, for example when $\Omega$ is convex and  $\gamma$ is a circle. But I've seen people claming this in many other contexts, like when $\Omega$ is an annulus around $0$ and $\gamma $ is an arbitrary curve inside it. This isn't obvious to me at all.  Thanks!