Integrating two exponentials produces a cosine integral? Can somebody explain? I discovered the following conversation that I do not understand. It reads:
$$\int_{-U_1}^0 {(\frac {u_1} {U_1}+1)e^{-j\omega_1u_1}}~du_1+\int_0^{U_1} {(-\frac {u_1} {U_1}+1)e^{-j\omega_1u_1}}~du_1 =
2 \int_0^{U_1} {(-\frac {u_1} {U_1}+1)cos(\omega_1u_1)}~du_1$$
How did they get from the two exponential integrals to the one with the cosine? Can somebody explain?
 A: You simply change the variable $v=-u_1$. Then you you can write the sum of your integrals as
$$
\int_0^{U_1}\left( -\frac{v}{U_1}+1 \right) \left( e^{-j\omega_1 v}+e^{j\omega_1 v} \right) 
$$
and if you remember that
$$
e^{jx}=\cos x +j\sin x
$$
you get your integral...
EDIT how to change the limits of integration.
Take your first integral 
$$
\int_{-U_1}^0 {(\frac {u_1} {U_1}+1)e^{-j\omega_1u_1}}~du_1 
$$
then put $v=-u_1$ and you get (remember you have $dv=-du_1$)
$$
-\int_{U_1}^0 {(\frac {-v} {U_1}+1)e^{-j\omega_1 (-v)}}~dv 
$$
then you can exchange the limit of the integral and change sign like this
$$
\int_a^bf(x)dx = -\int_b^a f(x)dx
$$
and therefore you get
$$
\int_{0}^{U_1} {(-\frac {v} {U_1}+1)e^{j\omega_1 v}}~dv 
$$
Is that clearer?
Let me know if is ok.
A: The first integral on the left side of the equation can be rewritten as
$$\int_{-U_1}^0 {(\frac {u_1} {U_1}+1)e^{-j\omega_1u_1}}~du_1=
\int_{0}^{U_1} {(-\frac {u_1} {U_1}+1)e^{j\omega_1u_1}}~du_1$$
and from
$$\cos x=\frac{1}{2}(e^{jx}+e^{-jx})$$
you obtain the result.
EDIT:
With $x=-u_1$ (and $dx=-du$) you get for the first integral on the left side of the equation
$$-\int_{U_1}^0 {(-\frac {x} {U_1}+1)e^{j\omega_1x}}~dx$$
The sign is changed by interchanging the lower and upper integration limits:
$$\int_{0}^{U_1} {(-\frac {x} {U_1}+1)e^{j\omega_1x}}~dx$$
