1. **State the problem:** Show that $$\int_0^\infty \frac{w \sin(wx)}{1+x^2} \, dw = \frac{\pi}{2} e^x$$ and $$\int_0^\infty \frac{\cos(wx)}{1+x^2} \, dw = \frac{\pi}{2} e^x$$ 2. **Recall the Fourier integral representation:** The Fourier sine and cosine transforms of functions involving exponential decay often yield expressions involving $e^x$. 3. **Step 1: Consider the integral involving sine:** We use the known integral formula: $$\int_0^\infty \frac{w \sin(aw)}{w^2 + b^2} \, dw = \frac{\pi}{2} e^{-ab}$$ for $a,b > 0$. 4. **Match the given integral:** Rewrite denominator $1+x^2$ as $w^2 + b^2$ with $b=1$ and variable $x$ as $a$. 5. **Apply the formula:** $$\int_0^\infty \frac{w \sin(wx)}{1 + w^2} \, dw = \frac{\pi}{2} e^{-x}$$ 6. **Check the sign in the exponent:** The problem states $e^x$ but the integral formula yields $e^{-x}$, so likely the problem has a typo or $x$ is negative. 7. **Similarly for cosine integral:** Use the formula: $$\int_0^\infty \frac{\cos(ax)}{x^2 + b^2} \, dx = \frac{\pi}{2b} e^{-a b}$$ 8. **Apply to given integral:** $$\int_0^\infty \frac{\cos(wx)}{1 + w^2} \, dw = \frac{\pi}{2} e^{-x}$$ 9. **Summary:** Both integrals equal $\frac{\pi}{2} e^{-x}$, assuming $x > 0$. **Final answers:** $$\int_0^\infty \frac{w \sin(wx)}{1 + w^2} \, dw = \frac{\pi}{2} e^{-x}$$ $$\int_0^\infty \frac{\cos(wx)}{1 + w^2} \, dw = \frac{\pi}{2} e^{-x}$$ This matches the Fourier integral representation results.