Math Problem Statement
If f(x)= [π/2 ,[x]<1
0 ,[x]>1] show that
F(x)= ∫0∞ cosλxsinλ/ λ d λ. Hence show that ∫0∞ sin λ/ λ d λ=π/2
Solution
Solution:
We start with the given piecewise definition of :
\begin{cases} \frac{\pi}{2} & \text{if } |x| < 1, \\ 0 & \text{if } |x| > 1. \end{cases}$$ We need to show that: $$F(x) = \int_0^\infty \frac{\cos(\lambda x) \sin \lambda}{\lambda} \, d\lambda.$$ --- #### Step 1: Simplify the Integral The given $$F(x)$$ can be rewritten by combining the trigonometric terms inside the integral: $$F(x) = \int_0^\infty \frac{\cos(\lambda x) \sin \lambda}{\lambda} \, d\lambda.$$ Using the product-to-sum formula: $$\cos(\lambda x)\sin(\lambda) = \frac{1}{2} \left[ \sin(\lambda(1+x)) - \sin(\lambda(1-x)) \right],$$ we rewrite the integral: $$F(x) = \frac{1}{2} \int_0^\infty \frac{\sin(\lambda(1+x))}{\lambda} \, d\lambda - \frac{1}{2} \int_0^\infty \frac{\sin(\lambda(1-x))}{\lambda} \, d\lambda.$$ --- #### Step 2: Connection to the Sine Integral and Convergence The integral $$\int_0^\infty \frac{\sin(a\lambda)}{\lambda} \, d\lambda,$$ is a well-known Fourier transform result and converges for $$|a| \leq 1$$. Its value is: $$\int_0^\infty \frac{\sin(a\lambda)}{\lambda} \, d\lambda = \frac{\pi}{2}, \quad \text{if } a = 1.$$ For our problem: - When $$|1+x| < 1$$, $$\int_0^\infty \frac{\sin(\lambda(1+x))}{\lambda} \, d\lambda = \frac{\pi}{2}$$. - Similarly, $$\int_0^\infty \frac{\sin(\lambda(1-x))}{\lambda} \, d\lambda = \frac{\pi}{2}$$ when $$|1-x| < 1$$. When $$|x| > 1$$, one or both of these integrals vanish because the sine function oscillates too rapidly, and their contributions cancel out. Thus, combining the results, we observe: $$F(x) = \begin{cases} \frac{\pi}{2} & \text{if } |x| < 1, \\ 0 & \text{if } |x| > 1. \end{cases}$$ This matches the given $$f(x)$$, proving $$F(x) = f(x)$$. --- #### Step 3: Showing $$\int_0^\infty \frac{\sin \lambda}{\lambda} \, d\lambda = \frac{\pi}{2}$$ To find $$\int_0^\infty \frac{\sin \lambda}{\lambda} \, d\lambda$$, we set $$x = 0$$ in $$F(x)$$. Substituting into $$F(x)$$, we have: $$F(0) = \int_0^\infty \frac{\cos(0) \sin \lambda}{\lambda} \, d\lambda = \int_0^\infty \frac{\sin \lambda}{\lambda} \, d\lambda.$$ From the definition of $$F(x)$$ and $$f(x)$$, we know $$F(0) = f(0) = \frac{\pi}{2}$$. Thus: $$\int_0^\infty \frac{\sin \lambda}{\lambda} \, d\lambda = \frac{\pi}{2}.$$ --- ### Final Answer: 1. $$F(x) = \int_0^\infty \frac{\cos(\lambda x) \sin \lambda}{\lambda} \, d\lambda$$ matches $$f(x)$$. 2. The integral $$\int_0^\infty \frac{\sin \lambda}{\lambda} \, d\lambda = \frac{\pi}{2}$$. --- ### Questions for Further Practice: 1. Derive $$\int_0^\infty \frac{\cos(\lambda) \sin(\lambda)}{\lambda} \, d\lambda$$ using a similar approach. 2. Prove the Fourier sine transform identity for $$\int_0^\infty \frac{\sin(\lambda a)}{\lambda} \, d\lambda$$. 3. Explore the convergence properties of $$\int_0^\infty \frac{\sin(\lambda)}{\lambda} \, d\lambda$$ for larger $$\lambda$$. 4. Verify $$\int_{-\infty}^\infty \frac{\sin \lambda}{\lambda} \, d\lambda = \pi$$. 5. Solve $$\int_0^\infty e^{-\lambda} \frac{\sin \lambda}{\lambda} \, d\lambda$$. --- ### Tip: The integral $$\int_0^\infty \frac{\sin(\lambda a)}{\lambda} \, d\lambda = \frac{\pi}{2}$$ for $$|a| = 1$$ is a crucial result in Fourier analysis.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Definite integrals
Fourier analysis
Trigonometric identities
Formulas
f(x) = [π/2, |x| < 1; 0, |x| > 1]
F(x) = ∫0∞ cos(λx)sin(λ)/λ dλ
cos(λx)sin(λ) = 1/2 [sin(λ(1+x)) - sin(λ(1-x))]
Theorems
Fourier transform properties
Integral convergence for oscillatory functions
Key result: ∫0∞ sin(λ)/λ dλ = π/2 for |a| = 1
Suitable Grade Level
Undergraduate level (Mathematics, Physics, or Engineering)
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