Integrating two exponentials produces a cosine integral? Can somebody explain?

Question

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?

Answer 1

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.

Answer 2

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:

$$\int_{0}^{U_1} {(-\frac {x} {U_1}+1)e^{j\omega_1x}}~dx$$