I've been trying to solve the improper integral of
ln(x)/sqrt(x) dx as a=1 and b=infinity
I am required to solve for this question and according to my answer, it is divergent. However, my answer is infinity-4. Is it acceptable to say that infinity minus a number is simply infinity? I just want to make sure whether I'm doing this right or not.
let u=sqrtx
du=1/2sqrtx dx
or dx=2sqrtx du=2u du
yes, inf-constant=inf or divergent
Thank you!
To solve the improper integral of ln(x)/sqrt(x) from 1 to infinity, you need to use the definition of an improper integral and evaluate the limit as the upper bound approaches infinity.
Let's go through the steps to solve the integral:
1. Start with the improper integral expression:
integral(ln(x)/sqrt(x), 1, infinity)
2. Apply L'Hospital's rule to evaluate the limit at the upper bound (infinity):
lim(x -> infinity) ln(x)/sqrt(x)
3. Rewrite ln(x) as 1/ln(e^x):
lim(x -> infinity) (1/(ln(e^x)/sqrt(x)))
4. Simplify the expression:
lim(x -> infinity) (sqrt(x)/ln(e^x))
5. Now, apply L'Hospital's rule again. Differentiate the numerator and denominator:
lim(x -> infinity) (1/((1/2)sqrt(x)/(1/x)))
6. Simplify further:
lim(x -> infinity) (x/(1/2)sqrt(x))
7. Divide the terms inside the limit by sqrt(x):
lim(x -> infinity) (2/(1/sqrt(x)))
8. Simplify to obtain:
lim(x -> infinity) 2sqrt(x)
Now, as x approaches infinity, the value of 2sqrt(x) also approaches infinity. Therefore, the integral is divergent.
Regarding your answer of "infinity-4", it is not correct to say that infinity minus a number is simply infinity. In mathematics, infinity is an abstract concept that does not follow ordinary arithmetic rules. Therefore, subtracting a number from infinity does not yield a meaningful result.
In conclusion, the improper integral of ln(x)/sqrt(x) from 1 to infinity is divergent.