I have many a times encountered (and used myself) the following technique:

$$\int \sin x \mathrm{d}x = \int \operatorname{Im}(e^{ix}) \mathrm{d}x = \operatorname{Im} \left( \int e^{ix} \mathrm{d}x \right) = \operatorname{Im}( -ie^{ix}) + C = -\cos x + C$$

Not only in this case, but I've used this kind of transform many a times, instinctively, to solve many of those monster trig integrals (and it works like a miracle) but never justified it.

Why and how is this interchange of integral and imaginary part justified?

At first, I thought it might be always true that we can do such a type of interchange anywhere, so, I tried the following: $\operatorname{Im}(f(z)) = f(\operatorname{Im}(z))$. But this is clearly not true, as the LHS is always real but RHS can be, possibly, complex too.

Second thoughts. I realized that we are dealing with operators here and not functions really. Both integral and imaginary parts are operators. So we have a composition of operators and we are willing to check when do these operators commute? I couldn't really make out any further conclusions from here and am stuck with the following questions:

When and why is the following true: $\int \operatorname{Im}(f(z)) \mathrm{d}z= \operatorname{Im} \left( \int f(z) \mathrm{d}z \right)$? (Provided that $f$ is integrable)

Is it always true? (Because like I've used it so many times and never found any counter example)

Edit : I am unfamiliar with integration of complex-valued functions but what I have in mind is that while doing such a thing, I tend to think of $i$ as just as some constant (Ah! I hope this doesn't sounds like really weird), as I stated in the example in the beginning. To be more precise, I have something of like this in my mind: because a complex-valued function $f(z)$ can be thought of as $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ where $u$ and $v$ are real-valued functions and we can now use our definition for integration of real-valued functions as $$\int f(z) \mathrm{d}z = \int (u(x,y) + iv(x,y)) \mathrm{d}(x+iy) = \left(\int u\mathrm{d}x - \int v\mathrm{d}y\right) +i\left(\int v\mathrm{d}x + \int u\mathrm{d}y\right)$$

  • 1
    $\begingroup$ What is your definition of $\int f(x) dx$ for a complex-valued function $f$? $\endgroup$ – Martin R Feb 11 '17 at 17:14
  • $\begingroup$ I have answered your question as an edit in the original post. $\endgroup$ – kishlaya Feb 11 '17 at 17:29

You can always write $f = \operatorname{Re}(f)+i\operatorname{Im}(f)$. Then, by linearity $\int f = \int \operatorname{Re}(f)+i\int \operatorname{Im}(f)$. But this is clearly the unique decomposition of $\int f$ in its real and imaginary part since both $\int \operatorname{Re}(f)$ and $\int \operatorname{Im}(f)$ are real numbers, hence we must have $\operatorname{Re}\int f = \int \operatorname{Re}f$ and the same for the imaginary part.

This is by the way a special case of the following more general observation:

If $E,F$ are complex Banach lattices and $T:E\to F$ is a real operator, i.e. mapping real elements to real elements, then $T\circ \operatorname{Re} = \operatorname{Re}\circ T$. Positive Operators are a special case of real operators and your question is a special case if we set $E = L^1, F=\mathbb C, T=\int$.

  • $\begingroup$ Oh! I got it. Wow, this was amazing. Btw, Banach lattices sound really interesting to me. Would you like to recommend me some references? $\endgroup$ – kishlaya Feb 11 '17 at 17:49
  • 2
    $\begingroup$ The classic book for Banach lattices is called "Banach lattices and positive operators" by Schaefer, but it is really hard to understand. I like the book "Positive Operators" by Charalambos D. Aliprantis and Owen Burkinshaw and the book "Banach lattices" by Peter Meyer-Nieberg. I found a short introduction here: siba-ese.unisalento.it/index.php/quadmat/article/viewFile/8711/… but I haven't read it, just had a short look. @kishlaya $\endgroup$ – Tim B. Feb 11 '17 at 18:16
  • $\begingroup$ @TimB., the result does not hold if the integral is a contour integral (i.e., with a $dz$ at the end and over a complex curve $\gamma$). This is because $\int_\gamma Re(f) dz$ can still be complex. $\endgroup$ – SuperM Aug 30 '19 at 17:52
  • 1
    $\begingroup$ @SuperM: It probably was not so clearly mentioned at the beginning but in the last part of the answer I state a more general result where one condition is that the Operator maps real elements to real elements. $\endgroup$ – Tim B. Oct 12 '19 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.