Hi, I needed help with this improper integral.
lower limit : 2
upper limit : infinity
integral of (lnx)/(x+1) dx
Thank You so much :D
To solve this improper integral, we'll use a common technique known as integration by parts. Here's how you can do it step by step:
Step 1: Split the Integral
Since the lower limit is a finite number, we can split the integral into two parts as follows:
∫[2,∞] (lnx)/(x+1) dx = ∫[2,∞] (lnx)dx - ∫[2,∞] (lnx)/(x+1) dx
Step 2: Evaluate the First Integral
Let's evaluate the first integral: ∫[2,∞] (lnx)dx
To solve this integral, we'll rewrite it using the property that ln(a) = ln|a|:
∫[2,∞] (lnx)dx = lim (t→∞) ∫[2,t] (lnx)dx
Now, we integrate using the power rule of integration for ln(x):
= lim (t→∞) [xlnx - x] from 2 to t
= lim (t→∞) [(tln t - t) - (2ln2 - 2)]
= lim (t→∞) (tln t - t) - (2ln2 - 2)
Since the limit of tln(t) as t approaches infinity is infinity and the limit of t as t approaches infinity is also infinity, we have an indeterminate form of ∞ - ∞. To resolve this, we can use L'Hôpital's rule:
= lim (t→∞) ((ln t + 1) / (1/t))
By applying L'Hôpital's rule to the above expression, we get:
= lim (t→∞) (1/t) / (-1/t^2)
= lim (t→∞) -t
= -∞
So, the first integral ∫[2,∞] (lnx)dx evaluates to -∞.
Step 3: Evaluate the Second Integral
We'll now evaluate the second integral: ∫[2,∞] (lnx)/(x+1) dx
To solve this integral, we'll use a substitution. Let u = lnx, which implies du = (1/x)dx.
When x = 2, u = ln(2). When x approaches infinity, u approaches infinity.
Substituting the above values, the integral becomes:
∫[ln(2),∞] (u)/(e^u + 1) du
This integral cannot be evaluated analytically; however, we can make some observations. As u increases without bound, the denominator e^u + 1 also increases without bound. In this case, the integral converges to a finite value called an improper integral. However, finding the exact value of this integral requires numerical approximations.
So, the second integral ∫[2,∞] (lnx)/(x+1) dx converges to a finite value.
Step 4: Combine the Results
Since the first integral is -∞ and the second integral converges to a finite value, we can write the overall result as:
∫[2,∞] (lnx)/(x+1) dx = -∞ + (Finite Value)
As a result, the improper integral diverges.