How do I show that real part of an integral equals the integral of the real part? $I=[a,b]$ with $a<b$ is compact range in $\mathbb R $. It is also given that $f,g \in  C(I)$ and $r,s\in I$. $C(I)$ is quantity of all in $I$ continuous functions. Show that:
$Re\int_{r}^{s} \! f(t) \, dt= \int_{r}^{s} \! Re f(t) \, dt$
and
$Im\int_{r}^{s} \! f(t) \, dt= \int_{r}^{s} \! Im f(t) \, dt$
With what should I start?
 A: It arises from the fact that the sum of real or imaginary parts is just the real or imaginary parts of the sum.
$$\sum{Re(z)}=Re\sum{z}$$
$$\sum{Im(z)}=Im\sum{z}$$
I implore you to test this identity by adding up a bunch of complex numbers and taking their real and imaginary parts.
The integral is nothing but an infinite sum. We will assume that our integral is Riemann Integrable. This would mean the following:
$$\int_a^b{f(x)}dx=\lim_{n\to \infty}\sum_{i=1}^{n}{f(x_i)\Delta x_i}$$
$$\frac{b-a}{n}=\Delta x_i$$
In this case our partitions are of equal intervals and constant and therefore $\Delta x_i$ is also constant.
The formula above would suggest that the following is true:
$$\int_a^b{Re(g(z))}dz=\lim_{n\to \infty}\sum_{i=1}^{n}{Re(g(z_i))\Delta z_i}$$
The identity on the very top tells us that the sum of the real part is just the real part of the sum.
$$=\lim_{n\to \infty}Re\sum_{i=1}^{n}{g(z_i)\Delta z_i}=Re\lim_{n\to \infty}\sum_{i=1}^{n}{g(z_i)\Delta z_i}=Re\int_a^b{g(z)}dz$$
And therefore:
$$\int_a^b{Re(g(z))}dz=Re\int_a^b{g(z)}dz$$
You will notice that the same is true for the imaginary part if you replace $Re(g(z))$ with $Im(g(z))$
A: If you have that the integral is additive, write $f=\text{Re}(f)+i\text{Im}(f)$. Then since both of Re and Im are continuous both functions are still integrable and $$\int f(x)dx=\int \text{Re}(f(x))dx+i\int \text{Im}(f(x))dx$$ take the real and imaginary parts of both sides to get the result.
