Vamos a resolver el problema paso a paso: --- ### 1. **Función dada** \[ f(x) = \begin{cases} 1, & -2 < x < 0 \\ -1, & 0 < x < 2 \\ 0, & \text{en otro caso} \end{cases} \] --- ### 2. **Transformada exponencial de Fourier** La transformada de Fourier está definida como: \[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx \] Para nuestra función, los intervalos donde \( f(x) \neq 0 \) son \( (-2, 0) \) y \( (0, 2) \): \[ F(k) = \int_{-2}^{0} 1 \cdot e^{-ikx} dx + \int_{0}^{2} (-1) \cdot e^{-ikx} dx \] Resolvamos cada parte: #### Primer integral: \[ \int_{-2}^{0} e^{-ikx} dx = \left[ \frac{e^{-ikx}}{-ik} \right]_{-2}^{0} = \frac{1}{-ik} \left( e^{-ik \cdot 0} - e^{-ik \cdot (-2)} \right) = \frac{1}{-ik} (1 - e^{2ik}) \] #### Segunda integral: \[ \int_{0}^{2} -e^{-ikx} dx = - \left[ \frac{e^{-ikx}}{-ik} \right]_{0}^{2} = \frac{1}{ik} \left( e^{-ik \cdot 2} - e^{-ik \cdot 0} \right) = \frac{1}{ik} (e^{-2ik} - 1) \] #### Sumando: \[ F(k) = \frac{1}{-ik}(1 - e^{2ik}) + \frac{1}{ik}(e^{-2ik} - 1) \] \[ = \frac{1 - e^{2ik} - (e^{-2ik} - 1)}{-ik} \] \[ = \frac{1 - e^{2ik} - e^{-2ik} + 1}{-ik} \] \[ = \frac{2 - e^{2ik} - e^{-2ik}}{-ik} \] \[ = \frac{2 - 2\cos(2k)}{-ik} \ \ \ \text{(usando } e^{ix} + e^{-ix} = 2 \cos x \text{)} \] \[ = \frac{2[1 - \cos(2k)]}{-ik} \] O también: \[ F(k) = \frac{2i [\cos(2k) - 1]}{k} \] --- ### 3. **Transformada seno de Fourier** La transformada seno de Fourier es: \[ S(k) = \int_{-\infty}^\infty f(x) \sin(kx) dx \] \[ S(k) = \int_{-2}^0 \sin(kx) dx + \int_0^2 -\sin(kx) dx \] \[ = \left[ -\frac{\cos(kx)}{k} \right]_{-2}^0 + \left[ \frac{\cos(kx)}{k} \right]_0^2 \] \[ = \left(-\frac{\cos(0)}{k} + \frac{\cos(-2k)}{k}\right) + \left(\frac{\cos(2k)}{k} - \frac{\cos(0)}{k}\right) \] Pero \(\cos(-2k) = \cos(2k)\) y \(\cos(0) = 1\): \[ = \left(-\frac{1}{k} + \frac{\cos(2k)}{k}\right) + \left(\frac{\cos(2k)}{k} - \frac{1}{k}\right) \] \[ = \frac{-1 + \cos(2k) + \cos(2k) - 1}{k} \] \[ = \frac{2\cos(2k) - 2}{k} \] \[ = \frac{2[\cos(2k) - 1]}{k} \] --- ### 4. **Escribir \( f(x) \) como una integral** Ya lo hicimos: \[ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(k) e^{ikx} dk \] --- ### 5. **Usar el resultado para evaluar la integral** El problema pide evaluar: \[ I = \int_0^\infty \frac{(\cos 2a - 1) \sin 2a}{a} da \] Nótese que en la transformada seno aparece \(\cos(2k) - 1\), así que la integral está relacionada. Observa que: \[ I = \frac{1}{2} \int_0^\infty \frac{2(\cos 2a - 1)\sin 2a}{a} da \] Vamos a usar la identidad: \[ 2(\cos 2a - 1)\sin 2a = 2\cos 2a \sin 2a - 2\sin 2a = \sin 4a - 2\sin 2a \] Por lo tanto: \[ I = \int_0^\infty \frac{\cos 2a - 1}{a} \sin 2a da = \frac{1}{2} \int_0^\infty \frac{\sin 4a - 2\sin 2a}{a} da \] Sabemos que: \[ \int_0^\infty \frac{\sin \alpha x}{x} dx = \frac{\pi}{2}, \quad \alpha > 0 \] Entonces: \[ I = \frac{1}{2} \left[ \int_0^\infty \frac{\sin 4a}{a} da - 2 \int_0^\infty \frac{\sin 2a}{a} da \right] \] \[ = \frac{1}{2} \left[ \frac{\pi}{2} - 2 \cdot \frac{\pi}{2} \right] \] \[ = \frac{1}{2} \left( \frac{\pi}{2} - \pi \right) \] \[ = \frac{1}{2} \left( -\frac{\pi}{2} \right) \] \[ = -\frac{\pi}{4} \] --- ## **Respuesta Final** \[ \boxed{ \int_0^\infty \frac{(\cos 2a - 1) \sin 2a}{a} da = -\frac{\pi}{4} } \] Si necesitas la respuesta justificada paso a paso en otro formato, dime.

Let's provide the solution step-by-step in English: --- ### Step 1: Understand the given function \(f(x)\) The function \(f(x)\) is defined as: \[ f(x) = \begin{cases} 1, & -2 < x < 0 \\ -1, & 0 < x < 2 \\ 0, & \text{elsewhere} \end{cases} \] --- ### Step 2: Find the Fourier exponential transform \(F(k)\) The Fourier transform is: \[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx \] Since \(f(x)\) is non-zero only between \(-2\) and \(2\): \[ F(k) = \int_{-2}^0 1 \cdot e^{-ikx} dx + \int_0^2 (-1) \cdot e^{-ikx} dx \] Compute each integral: - First integral: \[ \int_{-2}^0 e^{-ikx} dx = \left[\frac{e^{-ikx}}{-ik}\right]_{-2}^0 = \frac{1}{-ik} (e^{0} - e^{2ik}) = \frac{1 - e^{2ik}}{-ik} \] - Second integral: \[ \int_0^{2} -e^{-ikx} dx = - \left[\frac{e^{-ikx}}{-ik}\right]_0^{2} = \frac{1}{ik}(e^{-2ik} - 1) \] Adding both: \[ F(k) = \frac{1 - e^{2ik}}{-ik} + \frac{e^{-2ik} - 1}{ik} \] Rewrite: \[ F(k) = \frac{1 - e^{2ik}}{-ik} + \frac{e^{-2ik} - 1}{ik} \] Combine into a common expression: \[ F(k) = \frac{-(1 - e^{2ik})}{ik} + \frac{e^{-2ik} - 1}{ik} = \frac{-1 + e^{2ik} + e^{-2ik} - 1}{ik} \] Simplify numerator: \[ = \frac{e^{2ik} + e^{-2ik} - 2}{ik} \] Recall that: \[ e^{ix} + e^{-ix} = 2 \cos x \] So: \[ F(k) = \frac{2 \cos 2k - 2}{ik} = \frac{2(\cos 2k - 1)}{ik} \] --- ### Step 3: Find the sine Fourier transform \(S(k)\) The sine transform: \[ S(k) = \int_{-\infty}^{\infty} f(x) \sin(kx) dx \] Similarly, only the intervals contribute: \[ S(k) = \int_{-2}^0 \sin(kx) dx + \int_0^2 (-\sin(kx)) dx \] Compute each: - First integral: \[ \int_{-2}^0 \sin(kx) dx = \left[-\frac{\cos(kx)}{k}\right]_{-2}^0 = -\frac{\cos 0}{k} + \frac{\cos(-2k)}{k} = -\frac{1}{k} + \frac{\cos 2k}{k} \] - Second integral: \[ \int_0^{2} -\sin(kx) dx = - \left[ -\frac{\cos(kx)}{k} \right]_0^{2} = \frac{\cos 2k - 1}{k} \] Adding: \[ S(k) = \left(-\frac{1}{k} + \frac{\cos 2k}{k}\right) + \frac{\cos 2k - 1}{k} = \frac{-1 + \cos 2k + \cos 2k - 1}{k} = \frac{2 \cos 2k - 2}{k} \] --- ### Step 4: Express \(f(x)\) via the Fourier integral Using inverse Fourier transform: \[ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(k) e^{ikx} dk \] --- ### Step 5: Evaluate the integral involving \(\cos 2a - 1\) and \(\sin 2a\): Given the relation found in the Fourier analysis, the integral: \[ I = \int_0^\infty \frac{(\cos 2a - 1) \sin 2a}{a} da \] can be related to the Fourier transforms because: \[ 2(\cos 2a - 1) \sin 2a = \sin 4a - 2 \sin 2a \] This is a key trigonometric identity. Thus: \[ I = \frac{1}{2} \int_0^\infty \frac{\sin 4a - 2 \sin 2a}{a} da \] Recall the standard integral: \[ \int_0^\infty \frac{\sin \beta x}{x} dx = \frac{\pi}{2} \quad \text{for } \beta > 0 \] Applying this: \[ \int_0^\infty \frac{\sin 4a}{a} da = \frac{\pi}{2} \] and \[ \int_0^\infty \frac{\sin 2a}{a} da = \frac{\pi}{2} \] Therefore: \[ I = \frac{1}{2} \left( \frac{\pi}{2} - 2 \cdot \frac{\pi}{2} \right) = \frac{1}{2} \left( \frac{\pi}{2} - \pi \right) = \frac{1}{2} \left( -\frac{\pi}{2} \right) = -\frac{\pi}{4} \] --- ### **Final answer:** \[ \boxed{ \int_0^\infty \frac{(\cos 2a - 1) \sin 2a}{a} da = -\frac{\pi}{4} } \] This completes the step-by-step solution and the final result.