Evaluate `int_(0)^(oo)(dt)/(t)`

Answer

To evaluate the integral

\[ I = \int_{0}^{\infty} \frac{dt}{t}, \]

we need to analyze the behavior of the integrand \(\frac{1}{t}\) as \(t\) approaches the limits of integration.

### Step 1: Split the Integral
The integral from \(0\) to \(\infty\) can be split into two parts:

\[ I = \int_{0}^{1} \frac{dt}{t} + \int_{1}^{\infty} \frac{dt}{t}. \]

### Step 2: Evaluate the First Integral
Now, we evaluate the first integral:

\[ \int_{0}^{1} \frac{dt}{t}. \]

This integral is improper because \(\frac{1}{t}\) approaches infinity as \(t\) approaches \(0\). We can evaluate it using a limit:

\[ \int_{0}^{1} \frac{dt}{t} = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} \frac{dt}{t} = \lim_{\epsilon \to 0^+} [\ln(t)]_{\epsilon}^{1} = \lim_{\epsilon \to 0^+} (\ln(1) - \ln(\epsilon)). \]

Since \(\ln(1) = 0\) and \(\ln(\epsilon) \to -\infty\) as \(\epsilon \to 0^+\), we have:

\[ \int_{0}^{1} \frac{dt}{t} = -(-\infty) = \infty. \]

### Step 3: Evaluate the Second Integral
Next, we evaluate the second integral:

\[ \int_{1}^{\infty} \frac{dt}{t}. \]

This integral is also improper, and we can evaluate it using a limit:

\[ \int_{1}^{\infty} \frac{dt}{t} = \lim_{b \to \infty} \int_{1}^{b} \frac{dt}{t} = \lim_{b \to \infty} [\ln(t)]_{1}^{b} = \lim_{b \to \infty} (\ln(b) - \ln(1)). \]

Again, since \(\ln(1) = 0\) and \(\ln(b) \to \infty\) as \(b \to \infty\), we have:

\[ \int_{1}^{\infty} \frac{dt}{t} = \infty. \]

### Step 4: Combine the Results
Since both integrals diverge to infinity, we conclude that:

\[ I = \infty + \infty = \infty. \]

Thus, the value of the integral is:

\[ \int_{0}^{\infty} \frac{dt}{t} = \infty. \]

### Final Answer
The integral diverges:

\[ \int_{0}^{\infty} \frac{dt}{t} = \infty. \]