integral zero to infinity, 1/(e^6x + e^-6x) dx ; does it converge? and if it doe
ID: 3215447 • Letter: I
Question
integral zero to infinity, 1/(e^6x + e^-6x) dx ; does it converge? and if it does how does it solve?Explanation / Answer
Given ?(0 to 8) p(x) e^(-x) dx: Use integration by parts. u = p(x), dv = e^(-x) dx du = p'(x) dx, v = -e^(-x). Therefore, ?(0 to 8) p(x) e^(-x) dx = -p(x)/e^x {for x = 0 to 8} + ?(0 to 8) p'(x) e^(-x) dx By n applications of L'Hopital's Rule, lim(t-->8) p(t)/e^t = lim(t-->8) [constant]/e^t = 0. Thus, ?(0 to 8) p(x) e^(-x) dx = p(0) + ?(0 to 8) p'(x) e^(-x) dx. Now, we repeatedly apply this formula until the polynomial is constant. **Suppose p(x) is of degree n, so that p^(n)(x) is constant. Then, we obtain ?(0 to 8) p(x) e^(-x) dx = p(0) + ?(0 to 8) p'(x) e^(-x) dx = p(0) + [p'(0) + ?(0 to 8) p''(x) e^(-x) dx] = p(0) + p'(0) + [p''(0) + ?(0 to 8) p'''(x) e^(-x) dx] ... = p(0) + p'(0) + ... + [p^(n-1)(0) + ?(0 to 8) p^(n)(x) e^(-x) dx] = p(0) + p'(0) + ... + [p^(n-1)(0) - p^(n)(x) e^(-x) {for x = 0 to 8}], since p^(n) is constant = p(0) + p'(0) + ... + p^(n-1)(0) - (0 - p^(n)(0)) = p(0) + p'(0) + ... + p^(n-1)(0) + p^(n)(0). ----------- Now, we use this to evaluate ?(0 to 8) (x^5 - 3x^2 + 1) e^(-x) dx Here, p(x) = x^5 - 3x^2 + 1 ==> p(0) = 1 p'(x) = 5x^4 - 6x ==> p'(0) = 0 p''(x) = 20x^3 - 6 ==> p''(0) = -6 p'''(x) = 60x^2 ==> p'''(0) = 0 p''''(x) = 120x ==> p''''(0) = 0 p'''''(x) = 120 ==> p'''''(0) = 120 So, the integral equals 1 + 0 + (-6) + 0 + 0 + 120 = 115. (Verified without this formula with Wolfram Alpha!) ------------------- *For the next integrals, I display the comparison and leave the details of the (easier) integrations for the comparison test to you. a) Note that x/sqrt(1 + x^6) x^3/x^4 = 1/x for all x >= 2. Since ?(2 to 8) dx/x diverges (to 8), we have that ?(2 to 8) (x^3 + 2x + 1)dx/(x^4 - 1) also diverges by the Comparison Test. c) Note that sin(x) / sqrt(x) 0. Since ?(0 to p/2) dx/sqrt(x) converges (to 2 sqrt(p/2)), ?(0 to p/2) sin x dx/sqrt(x) also converges by the Comparison Test. ----------------------------Related Questions
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