Why does integrating with real parts work?

Ideas and Question

When you integrate with real parts can you exclude the imaginary part because it extends beyond the Cartesian plane? or is there something with only using the real part of the function? Maybe somewhere my math was wrong, but why does this work?

Example

Taking

1.)$∫e^x cos(x) dx$ , and since $$e^ix=cos(x)+isin(x)$$

2.) You can use Real$∫e^x*e^{ix} dx$= Real$∫e^{(1+i)x} dx$

3.) Substitute u= (1+i)x, with du=i+1 and dx= $\frac {1}{(1+i)}$

4.) $So, Real \frac{1}{1+i}∫e^udu] -> Real \frac{e^x*e^ix}{1+i}$

5.) Substituting cos(x)+isin(x) in for $e^{ix}$

6.) Real$\frac{e^x(cos(x)+isin(x)}{1+i}$, then multiplying $\frac{num}{denom}$ by the conjugate of (1+i)

7.) Real$\frac{e^x(cos(x)+e^x(isin(x)-ie^x(cos(x)+e^x(sin(x)}{2}$

8.) Which can seperate to Real$\frac{e^x(cos(x)+sin(x))}{2} + \frac{ie^xsin(x)-cos(x)}{2}$

9) Then, since it's the real part the answer is $\frac{e^x(cos(x)+sin(x)}{2} + c_1$

But why can you do this?

• Hi, welcome to MSE! You can format your equations with mathjax :) – Mauve Feb 26 '18 at 1:34
• thanks, I'll work on that now – user535189 Feb 26 '18 at 1:35
• A simplified explanation: an integral is roughly speaking a sum and the sum of real parts $\sum_i \Re [z_i]$ equals the real part of the sum $\Re[\sum_i z_i]$ (here the first sum is the real integral you want to find and the second sum is the real part of the complex integral) – Winther Feb 26 '18 at 1:50

In Step 2 of your method, using linearity of the integral, we have that

$$\int e^x e^{ix} dx$$ $$= \int e^x ( \cos(x) + i \sin(x) ) dx$$ $$= \int e^x \cos(x) + e^x i \sin(x) dx$$ $$= \int e^x \cos(x) dx + \int e^x i \sin(x) dx$$ $$= \int e^x \cos(x) dx + i \int e^x \sin(x) dx$$

now take the real part, which is what you want to integrate.

The long and short of it is "because that's the way it works." Read my answer for a slightly elaborated explanation.

In your example, you want to compute $$\int e^{x}\cos{x}\,\mathrm{d}x.$$ Your teacher has taught you a neat trick, which is to focus instead on the integral $$\int e^{(1+i)x}\,\mathrm{d}x.$$ Your teacher has pointed out to you that the integral you want to compute is just the real part of this, and you want to know why this "works."

Take the integral your teacher gave you and note that the integrand obeys the identity $e^{(1+i)x}=e^{x}e^{ix}=e^{x}(\cos{x}+i\sin{x}).$ So sub this in and multiply out: \begin{align*} \int e^{(1+i)x}\,\mathrm{d}x & = \int e^{x}(\cos{x}+i\sin{x})\,\mathrm{d}x\\ & = \int (e^{x}\cos{x}+ie^{x}\sin{x})\,\mathrm{d}x \end{align*} Now the big question: how do we actually compute an integral of a complex integrand? Here's the answer: when the integrand is a complex-valued function of a single real variable, we just split the integral into real and imaginary parts and integrate the two separately: $$\int (e^{x}\cos{x}+ie^{x}\sin{x})\,\mathrm{d}x = \int e^{x}\cos{x}\,\mathrm{d}x+i\int e^{x}\sin{x}\,\mathrm{d}x.$$ From this it is clear that the real part is what you want.

But wait! I hear you cry. Why are we allowed to split the integrand? Here's the bit you might not like: we are allowed to do it because that is the way we define complex integrals. It's true by fiat. We get to make the rules, and we decided that's what the rules should say. Why is that a good definition? I think the best explanation is to think of complex numbers as vectors and think about how we should integrate vector-valued functions of a real variable.

• Of course, this leaves open the question, "Why are we also allowed to evaluate the integral $\int e^{(1+i)x}\,\mathrm{d}x$ as $e^{(1+i)x}/(1+i)+C$?" That really does need justification, the details of which I doubt your teacher will want to get into. – Will R Feb 26 '18 at 11:13

At the time I was trained, we did not use directly real and imaginary parts . To show you how we used to proceed is illustrated below.

Let us consider $$I=\int e^x \cos(x)\,dx$$ and associate to it $$J=\int e^x \sin(x)\,dx$$ making $$I+iJ=\int e^{(1+i)x}\,dx=\frac{e^{(1+i)x} }{1+i }=\frac 12(1-i)e^{(1+i)x}$$ $$I-iJ=\int e^{(1-i)x}\,dx=\frac{e^{(1-i)x} }{1-i }=\frac 12(1+i)e^{(1-i)x}$$ Now, work with $$I=\frac{(I+iJ)+(I-iJ) }2$$ expanding the expontial and reusing $e^{ix}=\cos(x)+i \sin(x)$