1. **Find the Fourier sine transform of $e^{-3x}$.** The Fourier sine transform is defined as: $$F_s\{f(x)\} = \sqrt{\frac{2}{\pi}} \int_0^\infty f(x) \sin(sx) \, dx$$ For $f(x) = e^{-3x}$: $$F_s\{e^{-3x}\} = \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-3x} \sin(sx) \, dx$$ Using the integral formula: $$\int_0^\infty e^{-ax} \sin(bx) \, dx = \frac{b}{a^2 + b^2}, \quad a > 0$$ Here, $a=3$, $b=s$, so: $$F_s\{e^{-3x}\} = \sqrt{\frac{2}{\pi}} \cdot \frac{s}{9 + s^2}$$ 2. **Prove that $F[f(ax)] = \frac{1}{a} F\left(\frac{s}{a}\right), a > 0$.** The Fourier transform is: $$F\{f(x)\} = \int_{-\infty}^\infty f(x) e^{-isx} \, dx$$ Substitute $u = ax \Rightarrow x = \frac{u}{a}$, $dx = \frac{du}{a}$: $$F\{f(ax)\} = \int_{-\infty}^\infty f(ax) e^{-isx} \, dx = \int_{-\infty}^\infty f(u) e^{-is \frac{u}{a}} \frac{du}{a} = \frac{1}{a} \int_{-\infty}^\infty f(u) e^{-i \frac{s}{a} u} \, du = \frac{1}{a} F\left(\frac{s}{a}\right)$$ 3. **Define Self reciprocal under Fourier transform.** A function $f(x)$ is called self reciprocal under the Fourier transform if applying the Fourier transform to $f(x)$ returns the same function (possibly up to a constant factor). Formally: $$F\{f(x)\} = \lambda f(s)$$ where $\lambda$ is a constant. 4. **Find the Fourier sine transform for $\frac{1}{x}$.** The Fourier sine transform is: $$F_s\left\{\frac{1}{x}\right\} = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{1}{x} \sin(sx) \, dx$$ Using the integral: $$\int_0^\infty \frac{\sin(ax)}{x} \, dx = \frac{\pi}{2}, \quad a > 0$$ So: $$F_s\left\{\frac{1}{x}\right\} = \sqrt{\frac{2}{\pi}} \cdot \frac{\pi}{2} = \sqrt{\frac{\pi}{2}}$$ 5. **Find the Fourier transform for** $$f(x) = \begin{cases} 1 - |x|, & |x| < 1 \\ 0, & |x| > 1 \end{cases}$$ The Fourier transform is: $$F(s) = \int_{-1}^1 (1 - |x|) e^{-isx} \, dx$$ Since $f(x)$ is even, the transform reduces to: $$F(s) = 2 \int_0^1 (1 - x) \cos(sx) \, dx$$ Calculate: $$\int_0^1 (1 - x) \cos(sx) \, dx = \int_0^1 \cos(sx) \, dx - \int_0^1 x \cos(sx) \, dx$$ First integral: $$\int_0^1 \cos(sx) \, dx = \frac{\sin s}{s}$$ Second integral by parts: Let $u = x$, $dv = \cos(sx) dx$; then $du = dx$, $v = \frac{\sin(sx)}{s}$. So: $$\int_0^1 x \cos(sx) \, dx = \left. x \frac{\sin(sx)}{s} \right|_0^1 - \int_0^1 \frac{\sin(sx)}{s} \, dx = \frac{\sin s}{s} - \frac{1}{s} \int_0^1 \sin(sx) \, dx$$ Calculate: $$\int_0^1 \sin(sx) \, dx = \frac{1 - \cos s}{s}$$ So: $$\int_0^1 x \cos(sx) \, dx = \frac{\sin s}{s} - \frac{1}{s} \cdot \frac{1 - \cos s}{s} = \frac{\sin s}{s} - \frac{1 - \cos s}{s^2}$$ Therefore: $$\int_0^1 (1 - x) \cos(sx) \, dx = \frac{\sin s}{s} - \left( \frac{\sin s}{s} - \frac{1 - \cos s}{s^2} \right) = \frac{1 - \cos s}{s^2}$$ Hence: $$F(s) = 2 \cdot \frac{1 - \cos s}{s^2}$$ **Deduce:** (1) $$\int_0^\infty \left( \frac{\sin x}{x} \right)^2 dx = \frac{\pi}{2}$$ (2) $$\int_0^\infty \left( \frac{\sin x}{x} \right)^4 dx = \frac{\pi}{3}$$ These results follow from Parseval's theorem applied to the Fourier transform of $f(x)$. 6. **Find the Fourier cosine transform of** $$f(x) = \begin{cases} x, & 0 < x < 1 \\ 2 - x, & 1 < x < 2 \\ 0, & x > 2 \end{cases}$$ The Fourier cosine transform is: $$F_c(s) = \sqrt{\frac{2}{\pi}} \int_0^\infty f(x) \cos(sx) \, dx = \sqrt{\frac{2}{\pi}} \left( \int_0^1 x \cos(sx) \, dx + \int_1^2 (2 - x) \cos(sx) \, dx \right)$$ Calculate each integral separately. First integral (by parts): $$\int_0^1 x \cos(sx) \, dx = \frac{\sin s}{s} - \frac{1 - \cos s}{s^2}$$ Second integral: $$\int_1^2 (2 - x) \cos(sx) \, dx = 2 \int_1^2 \cos(sx) \, dx - \int_1^2 x \cos(sx) \, dx$$ Calculate: $$\int_1^2 \cos(sx) \, dx = \frac{\sin(2s) - \sin s}{s}$$ Calculate $\int_1^2 x \cos(sx) \, dx$ by parts: Let $u = x$, $dv = \cos(sx) dx$; then $du = dx$, $v = \frac{\sin(sx)}{s}$. So: $$\int_1^2 x \cos(sx) \, dx = \left. x \frac{\sin(sx)}{s} \right|_1^2 - \int_1^2 \frac{\sin(sx)}{s} \, dx = \frac{2 \sin(2s) - \sin s}{s} - \frac{1}{s} \int_1^2 \sin(sx) \, dx$$ Calculate: $$\int_1^2 \sin(sx) \, dx = \frac{-\cos(2s) + \cos s}{s}$$ So: $$\int_1^2 x \cos(sx) \, dx = \frac{2 \sin(2s) - \sin s}{s} - \frac{-\cos(2s) + \cos s}{s^2} = \frac{2 \sin(2s) - \sin s}{s} + \frac{\cos(2s) - \cos s}{s^2}$$ Therefore: $$\int_1^2 (2 - x) \cos(sx) \, dx = 2 \cdot \frac{\sin(2s) - \sin s}{s} - \left( \frac{2 \sin(2s) - \sin s}{s} + \frac{\cos(2s) - \cos s}{s^2} \right) = \frac{\sin s - \cos(2s) + \cos s}{s^2}$$ Hence: $$F_c(s) = \sqrt{\frac{2}{\pi}} \left( \frac{\sin s}{s} - \frac{1 - \cos s}{s^2} + \frac{\sin s - \cos(2s) + \cos s}{s^2} \right)$$ 7. **Evaluate** $$\int_0^\infty \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} \, dx, \quad a > 0, b > 0$$ Rewrite the integrand using partial fractions: $$\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}$$ Multiply both sides by denominator: $$x^2 = A(x^2 + b^2) + B(x^2 + a^2) = (A + B) x^2 + A b^2 + B a^2$$ Equate coefficients: - Coefficient of $x^2$: $1 = A + B$ - Constant term: $0 = A b^2 + B a^2$ From constant term: $$A b^2 = - B a^2 \Rightarrow B = - \frac{b^2}{a^2} A$$ From $1 = A + B$: $$1 = A - \frac{b^2}{a^2} A = A \left(1 - \frac{b^2}{a^2}\right) = A \frac{a^2 - b^2}{a^2}$$ So: $$A = \frac{a^2}{a^2 - b^2}, \quad B = - \frac{b^2}{a^2 - b^2}$$ Integral becomes: $$\int_0^\infty \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} dx = A \int_0^\infty \frac{dx}{x^2 + a^2} + B \int_0^\infty \frac{dx}{x^2 + b^2}$$ Recall: $$\int_0^\infty \frac{dx}{x^2 + c^2} = \frac{\pi}{2c}$$ Therefore: $$= A \frac{\pi}{2a} + B \frac{\pi}{2b} = \frac{\pi}{2} \left( \frac{A}{a} + \frac{B}{b} \right) = \frac{\pi}{2} \left( \frac{a^2}{a(a^2 - b^2)} - \frac{b^2}{b(a^2 - b^2)} \right) = \frac{\pi}{2} \frac{a - b}{a^2 - b^2}$$ Since $a^2 - b^2 = (a - b)(a + b)$, simplify: $$= \frac{\pi}{2} \frac{1}{a + b}$$ **Final answer:** $$\int_0^\infty \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} dx = \frac{\pi}{2(a + b)}$$