Answer:
Step-by-step explanation:
Find the values of p for which the following integral converges:
∫[infinity]e 1/(x(ln(x))^p)dx
Input youranswer by writing it as an interval. Enter brackets or parentheses in the first and fourth blanks as appropriate, and enter the interval endpoints in the second and third blanks. Use INF and NINF (in upper-case letters) for positive and negative infinity if needed. If the improper integral diverges for all p, type an upper-case "D" in every blank.
Values of p are in the interval ,
For the values of p at which the integral converges, evaluate it. Integral =
Recall that
[tex]\int\limits^{\infty}_1 \frac{1}{x^p} dx[/tex]
converge if p > 1 and converge to [tex]\frac{1}{p-1}[/tex] and divertgent if p ≤ 1
Now, let u = Inx ⇒ du = 1/x dx
: e ≤ x ≤ ∞ ⇒ 1 ≤ u < ∞
⇒ [tex]\int\limits^{\infty}_e \frac{dx}{x(Inx)^p} = \int\limits^{\infty}_1 {x} \frac{du}{u^p}[/tex]
converge if p > 1 and converge to [tex]\frac{1}{p-1}[/tex] and divertgent if p ≤ 1
[tex]\text {Integral}=\frac{1}{p-1}[/tex]
For the values of p the integral converges as (1 / (P - 1)) and this can be determined by using the substitution method.
Given :
To solve the given integral substitute lnx as given below:
u = lnx
Differentiate the above expression.
[tex]du = \dfrac{1}{x}dx[/tex]
x du = dx
where [tex]\rm e \leq x \leq \infty \to 1\leq u\leq \infty[/tex].
Now, the given integral becomes:
[tex]\rm \int\limits^{\infty}_1 {\dfrac{1}{x(lnx)^p}} \, dx = \int\limits^{\infty}_1 {x} \dfrac{1}{u^P}\, du[/tex]
According to the given data, the value of 'p' lies in the interval (0,[tex]\infty[/tex]).
So, the above integral becomes:
[tex]\rm \int\limits^{\infty}_1 {\dfrac{1}{x(lnx)^p}} \, dx = \dfrac{1}{P-1}[/tex]
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https://brainly.com/question/18651211
