Why is treating $i$ as a constant in integration, valid?

Question

Why do we, when doing integrals like $\int i\cos xdx$, treat $i$ to be a constant? Is there any proof? Wolfram gives the answer simply as $i\sin x+\text{[constant]}$.
I have a confusion, because integration is just the reverse of finding derivatives, i.e, Let $f(x)$ be a function, in differentiation, we find $f'(x)$, but in integration, we find $F(x)$, where $F$ is the antiderivative of $f$. But that is in the real plane (Cartesian co-ordinates). Definition of derivatives also is defined in real plane. But then how can we do that in complex plane, when it is not defined for it.
Am I missing something? Please clear my doubts.
Thank you in advance.

$i$ is treated as a constant in integration because it's value does not change, or in other words value of $i$ does not deepened on any variable. — jimjim, Sep 16 '15 at 07:27
Okay, but don't you agree that derivative of a function is defined in real plane.? http://www.zweigmedia.com/RealWorld/Calcsumm3b.html. Then how can we treat $i$ as a constant? How is is at all defined? How is $\iota$ integrable and differentiable? — Aditya Agarwal, Sep 16 '15 at 07:30
Actually, the integrand function may be a function from real numbers to complex numbers. This means that $x$ may assume only real values, but that the final value is a complex number. If x is real, $x \mapsto x+ix$ a function of this kind. In any case, a complex derivative can be defined: check Holomorphic functions. — Lonidard, Sep 16 '15 at 07:31
There is no such thing as real plane, there is real line and complex plane. BTW it is not the introduction of i that makes it strange, i was always there, only it had a coefficient 0. — jimjim, Sep 16 '15 at 07:35
Of course. Accept the fact that $i$ is not a function of $x$ and that it is constant; from there, linearity of derivative and integral show that the computation is as easy as it seems! @Arjang $\mathbb{R}^2$ is usually referred to as "the real plane"; it is also isomorphic to $\mathbb{C}$ which is useful for analogies — Lonidard, Sep 16 '15 at 07:35
The elementary approach to this problem is to simply realize that $i$ is a constant, not a variable. Integrating $5 + 4i$ is the same as integrating $5$. That is one of the beautiful things about the proof of $c \int f(x) dx= \int f(x)cdx$ Now, proving that this is always so is tough if you require complex analysis, and knowledge of the Cauchy integral theorem is required for the proof (which is not very elementary) — Brevan Ellefsen, Sep 16 '15 at 07:37
@bharb : There is no need to accept anything without proof. i is not a constant on the real line, but it is a constant on complex plane. — jimjim, Sep 16 '15 at 07:37
"Cauchy Integral Theorem"? Eh, @BrevanEllefsen ? That link seems to pay off? — Aditya Agarwal, Sep 16 '15 at 07:37
@Arjang says that there is no need to accept anything without proof. I agree, there has to be a way without using non-elementary approach. — Aditya Agarwal, Sep 16 '15 at 07:39
@Arjang $i$ is not part of the real line. How can you even say it's constant-nonconstant? — Lonidard, Sep 16 '15 at 07:39
@AdityaAgarwal It wasn't a new concept, I've known about it... but it seemed like a good reference to make since you just got it. That doesn't mean I've ever bothered to read the proof... again, it's non elementary, I've started some papers on it in the past. — Brevan Ellefsen, Sep 16 '15 at 07:39
I would say the whole question here is defining what integration over a complex field actually means, and how it works. We should direct the focus there to avoid bickering in the commend section as we have been. — Brevan Ellefsen, Sep 16 '15 at 07:41
Elementary, is school-level approach. Not using abstract theories, simple, proof. — Aditya Agarwal, Sep 16 '15 at 07:45
The easiest definition I can find for complex derivatives is found on this page: http://mathfaculty.fullerton.edu/mathews/c2003/AnalyticFunctionMod.html It requires reworking the definition of the derivative to work. It should be noted that not all calculus rules derived for real functions carry to complex functions, such as the Mean Value Theorem. Everything else I can find requires knowledge of Contour integrals, which are not grade school level (as contours are not grade-school level) — Brevan Ellefsen, Sep 16 '15 at 07:47
Can't we prove that? @BrevanEllefsen. Because I don't seem to understand things without a proof — Aditya Agarwal, Sep 16 '15 at 07:48
I agree that it doesn't seem right that the page omits a proof and says "let's use our imagination"... of course, this may be to avoid dealing with complex analysis, but I'll keep looking for an elementary proof of the definition of the complex derivative. If we can find that then we have your elementary proof. — Brevan Ellefsen, Sep 16 '15 at 07:51
As far as I can tell (and anyone should correct me if I'm mistaken) but it's very hard to conceptualize the definition of a derivative in the complex plane due to the nature of complex numbers being composed of two parts... tangent lines become more confusing. However, functions which output complex numbers can still follow patterns... Take for example $f(x) = 5xi + 2$. Since just $x$ is a variable in that equation, we can view this as a function where the slope for any value of $x$ is $2i$. — Brevan Ellefsen, Sep 16 '15 at 08:07
$2i$ or $5i$? BTW, yes you got me.Take for example, $2\iota x=y$. Then $y'=2\iota$ Now take $g(x)=3\iota x$ and $g'(x)=3\iota$. Now who's slope is greater? — Aditya Agarwal, Sep 16 '15 at 08:10
Since $i$ is a constant in the complex plane, we basically go through the same process that gives us the real definition of the derivative, except we don't have a "change in y"... instead we have to define $z_0$, a point on the complex plane that is "holomorphic at the point", meaning the function is infinitely differentiable in the neighborhood of $z_0$. In essence, we want the limit as we approach $z_0$ to be the same from any direction we come from. (And yes, I meant $5i$, I just cant edit it now) — Brevan Ellefsen, Sep 16 '15 at 08:13
In essence, we want the limit as we approach z0 to be the same from any direction we come from". This sounds like finding a multivariable limit. — Aditya Agarwal, Sep 16 '15 at 08:15
More or less, yes. Look at this pic... https://upload.wikimedia.org/wikipedia/commons/thumb/c/c3/Non-holomorphic_complex_conjugate.svg/440px-Non-holomorphic_complex_conjugate.svg.png. It shows a function that isn't complex-differentiable only because we get different limits as we approach $z_0$ from different directions. The complex plane is defined to be two-dimensional, and so we could expect limits with it to act very similar in some aspects to multidimensional limits. — Brevan Ellefsen, Sep 16 '15 at 08:19
@bharb : for something to be constant/non-constant, it's value has to be independent/dependent. There are many constants that are not on real line : e.g. the matrix to rotate the plane by 90 degrees is not on the real line either, but it is constant matrix that rotates the plane by 90 degrees. To see if something is constant, if it's value does not depend on anything else then it is constant. For example $\pi$ is a value that is not constant and is on real line refer to : http://math.stackexchange.com/questions/17366/why-is-pi-equal-to-3-14159/17384#17384 — jimjim, Sep 16 '15 at 10:42
@AdityaAgarwal : $\pi$ depends on geometry, on sphere $\pi$ is not unique. Again that depends on how one defines $\pi$, what is the ratio of circumference to it's diameter for a circle on a sphere? If $\pi$ is defined to be the ratio of circumference to it's diameter for a circle, then it is different on plane and on a sphere, hence by that definition $\pi$ is not constant. How ever if it is suppose to be only circumference to it's diameter for a circle on the flat surface ( non-Euclidean geometry anyone?) then yes it is consntant, it just depends on how it is definied. — jimjim, Sep 16 '15 at 10:50

score 4 · Accepted Answer · answered Sep 16 '15 at 08:21

4

One can use fancier terms, but what you want to do is to extend integration (and differentiation) from real-valued functions to complex-valued functions. It is natural to do so in a way that preserves linearity (but with linearity over $\mathbf C$ instead of $\mathbf R$).

Every complex-valued function (of a real variable $x$) can be written as $$ f(x)=g(x)+ih(x) $$ where $g$ and $h$ are real valued. Assuming that we have defined differentiation for real-valued functions only, it is natural to define $$ f'(x)=g'(x)+ih'(x). $$ A consequence is that $$ \int g(x)+ih(x)\,dx=\int g(x)\,dx+i\int h(x)\,dx. $$ Luckily, this fits very well with all rules of calculations you are familiar with, and maybe that is why it is confusing. It also fits when you do definite integrals, since also there, you can divide into real and imaginary part and do Riemann sums on each part...

Finally, your integral $\int i\cos x\,dx$ is therefore $$ \int i\cos x\,dx=i\int \cos x\,dx=i\sin x+C. $$

answered Sep 16 '15 at 08:21

mickep

19,962

That doesn't answer my questions, sorry. But you just wrote my query mathematically. In the second last step, how can you take $i$ to be constant? *That is my question.* – Aditya Agarwal Sep 16 '15 at 08:24
@AdityaAgarwal I'm still not quite sure where you are getting at... think about it this way. We take it for granted that the number 5 is a constant, and is part of the real numbers. That's all fine and dandy, and we don't even have a concept of $i$ to deal with in the real numbers. When we move to complex numbers, $i$ is by definition $\sqrt{-1}$. Notice that -1 is a constant in our real numbers... we accept that. Also notice that the square root of a constant is still a constant. Therefore, $i$ must be a constant. Is this more what you're wanting? – Brevan Ellefsen Sep 16 '15 at 08:27
What do you mean by second last step? $\int i\cos x,dx=i\int \cos x,dx$? This follows from what is written above with $g(x)=0$ and $h(x)=\cos x$. – mickep Sep 16 '15 at 08:30
My question, is that how can we just use our imagination and extend integration and differentiation to complex numbers, when we can't even compare the slopes of two lines? – Aditya Agarwal Sep 16 '15 at 08:30
But we can do $\int af(x)dx=a\int f(x)dx$ only if $a$ is a constant. But my question here is why do we take $i$ to be a constant? Now, can you see the circularity? – Aditya Agarwal Sep 16 '15 at 08:31
1

@Aditya Agarwal If you have $\int i f(x)dx$, where $f(x)$ is real, then this integral "sits" on the imaginary axis, and since integration is linear we can rewrite this as $i\int f(x)dx$. This still "sits" on the imaginary axis. So there is no confusion between "imaginary or real slopes" at all. $i$ is a constant. – pshmath0 Sep 16 '15 at 08:32
What is an imaginary slope? (Intuitively) – Aditya Agarwal Sep 16 '15 at 08:33
Before we extended integration to complex-valued functions, we could only say that $\int a f(x),dx=a\int f(x),dx$ for real constants $a$. After the extension, the same property follows for complex constants $a$. I see no contradiction in that. – mickep Sep 16 '15 at 08:33
How can we extend integration over that without any proof? – Aditya Agarwal Sep 16 '15 at 08:34
Also, why do we have to be able to compare slopes of lines to be able to integrate? – mickep Sep 16 '15 at 08:34
Why not? Suppose I ask you what is $$\lim\limits_{x\to0}\frac{\int^x_0 i\cos\theta d\theta}{\int^x_0 \cos\theta d\theta}$$? Then? – Aditya Agarwal Sep 16 '15 at 08:35
The right-hand side equals $i$ (by complex linearity) for all $x\neq 0$, and hence the limit is $i$. – mickep Sep 16 '15 at 08:37
Now what is complex linearity? – Aditya Agarwal Sep 16 '15 at 08:38
I would say we think of imaginary slope as such: We have two parts to complex numbers, so they could change individually. If just the part corresponding to $i$ is changing, then we have an imaginary change in slope. If the real part changes, then we have a real change in slope. To get this all into one formula we have to combine both into one by making sure the limits match up for a point $z_0$, which allows us to view the whole complex number as one point. – Brevan Ellefsen Sep 16 '15 at 08:38
Complex linearity is $\int a f(x),dx=a\int f(x),dx$ for $a\in\mathbf C$ (and $\int f(x)+g(x),dx=\int f(x),dx+\int g(x),dx$). – mickep Sep 16 '15 at 08:39
Geometrically, what would this mean, (to the eye)? @BrevanEllefsen. Would there be any change in the visibility if we change the real part? – Aditya Agarwal Sep 16 '15 at 08:39
@AdityaAgarwal Here is a nice video I found for visualizing the concept https://www.youtube.com/watch?v=1rVHLZm5Aho – Brevan Ellefsen Sep 16 '15 at 08:44

Why is treating $i$ as a constant in integration, valid?

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