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
We aim to show that is related to the integral expression for and use it to deduce the value of the improper integral .
Step 1: Define the given function
The function is defined as:
\begin{cases} \frac{\pi}{2}, & |x| < 1, \\ 0, & |x| \geq 1. \end{cases}$$ ### Step 2: Define $$F(x)$$ and relate it to $$f(x)$$ The function $$F(x)$$ is given as: $$F(x) = \int_0^\infty \frac{\cos(\lambda x) \sin\lambda}{\lambda} \, d\lambda.$$ Using the trigonometric identity $$\cos(\lambda x) \sin\lambda = \frac{1}{2} \left[\sin(\lambda (x + 1)) - \sin(\lambda (x - 1))\right]$$, this becomes: $$F(x) = \frac{1}{2} \int_0^\infty \frac{\sin(\lambda (x + 1))}{\lambda} \, d\lambda - \frac{1}{2} \int_0^\infty \frac{\sin(\lambda (x - 1))}{\lambda} \, d\lambda.$$ ### Step 3: Evaluate the integrals The integral $$\int_0^\infty \frac{\sin(a\lambda)}{\lambda} \, d\lambda$$ for a real $$a > 0$$ is known to converge to $$\frac{\pi}{2}$$ if $$a = 1$$, and $$0$$ otherwise. #### Case $$|x| < 1$$: - If $$|x| < 1$$, $$x+1 > 0$$ and $$x-1 > -1$$, so the contributions from both integrals exist and simplify to: $$F(x) = \frac{\pi}{2}.$$ #### Case $$|x| \geq 1$$: - If $$|x| \geq 1$$, either $$x+1 > 1$$ or $$x-1 \geq 1$$, leading to cancellation in the sine terms. Hence: $$F(x) = 0.$$ Thus, $$F(x) = f(x)$$, as required. ### Step 4: Evaluate $$\int_0^\infty \frac{\sin\lambda}{\lambda} \, d\lambda$$ To evaluate $$\int_0^\infty \frac{\sin\lambda}{\lambda} \, d\lambda$$, note that $$F(0)$$ is given by: $$F(0) = \int_0^\infty \frac{\cos(0) \sin\lambda}{\lambda} \, d\lambda = \int_0^\infty \frac{\sin\lambda}{\lambda} \, d\lambda.$$ From the earlier result, $$F(0) = f(0) = \frac{\pi}{2}$$. Hence: \[ \int_0^\infty \frac{\sin\Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Improper Integrals
Trigonometric Identities
Fourier Analysis
Formulas
f(x) = {π/2, |x| < 1; 0, |x| ≥ 1}
F(x) = ∫₀^∞ (cos(λx) sinλ) / λ dλ
cos(λx)sinλ = 1/2 [sin(λ(x+1)) - sin(λ(x-1))]
∫₀^∞ (sin(λ)) / λ dλ = π/2
Theorems
Improper Integral Convergence
Sine and Cosine Trigonometric Identities
Suitable Grade Level
Undergraduate Mathematics
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