how do you solve the integral of 1/[(square root of x)(lnx)] from 2 to infinity?
i did the p- integral theorem with 1/square root of x and got it to be a divergent integral. however i was told this was the wrong way and that i should do it by integration by parts. but i can't figure it out by that method. please help. thanks.
You can proceed as follows. Substitute:
x = e^t. Then the integral becomes:
Integral from ln(2) to infinity of
e^(t/2)/t dt
We can get rid of the factor 2 in the exponential by putting y = t/2. The integral becomes:
Integral from 1/2 ln(2) to infinity of
e^(y)/y dy
For positive y we have
e^(y) > 1
It follows from this that the integral is larger than
Integral from 1/2 ln(2) to infinity of
1/y dy
but this is already divergent, so the integral diverges.
To solve the integral ∫(1/[(√x)(lnx)]) dx from 2 to infinity, you are correct that using the p-integral test would lead to a divergent result.
Instead, let's approach it using integration by parts.
Integration by parts is a technique that allows you to rewrite an integral involving two functions as a new integral that may be easier to solve. The formula for integration by parts is:
∫u dv = uv - ∫v du
In this case, we let u = ln(x) and dv = (1/√x) dx.
Now, let's find du and v:
dv = (1/√x) dx
Integrating dv, we find:
v = 2√x
To find du, differentiate u with respect to x:
du = (1/x) dx
Now we have all the necessary components to apply the integration by parts formula:
∫(1/[(√x)(lnx)]) dx = uv - ∫v du
Using the given values for u, v, du, and v, we can rewrite the equation as:
∫(1/[(√x)(lnx)]) dx = ln(x) * 2√x - ∫2√x * (1/x) dx
Simplifying further, we have:
∫(1/[(√x)(lnx)]) dx = 2√x * ln(x) - 2∫√x dx
To solve the remaining integral, we use a substitution. Let's substitute u = √x:
Now, we have ∫(1/[(√x)(lnx)]) dx = 2√x * ln(x) - 2∫2√x dx = 2√x * ln(x) - 4∫du
Integrating, we find:
∫(1/[(√x)(lnx)]) dx = 2√x * ln(x) - 4u + C
Remember that u = √x, so we substitute back in:
∫(1/[(√x)(lnx)]) dx = 2√x * ln(x) - 4√x + C
Now, let's evaluate the integral from 2 to infinity:
∫[2,∞] (1/[(√x)(lnx)]) dx = lim(h→∞) (∫[2,h] (1/[(√x)(lnx)]) dx)
Evaluating the integral, we get:
= lim(h→∞) (2√h * ln(h) - 4√h) - (2√2 * ln(2) - 4√2)
Taking the limit as h approaches infinity, the first term goes to infinity, and the second term remains constant. Therefore, the integral is divergent.
In conclusion, using the integration by parts method, we have determined that the integral ∫(1/[(√x)(lnx)]) dx from 2 to infinity is divergent.