`int_0^oo[3/(x^2+1)]dx,` where `[.]` denotes the greatest integer function, is equal to (A) `sqrt2` (B) `sqrt2+1` (C) `3/sqrt2` (D) infinite

Options

A. `sqrt2`
B. `sqrt2+1`
C. `3//sqrt2`
D. infinte

Answer

To solve the integral \( \int_0^\infty \left\lfloor \frac{3}{x^2 + 1} \right\rfloor dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we will analyze the behavior of the function \( \frac{3}{x^2 + 1} \) over the interval from \( 0 \) to \( \infty \).

### Step 1: Determine the range of the function \( \frac{3}{x^2 + 1} \)

The function \( \frac{3}{x^2 + 1} \) is continuous and decreases as \( x \) increases.

- At \( x = 0 \):
\[ \frac{3}{0^2 + 1} = 3 \]

- As \( x \to \infty \):
\[ \frac{3}{x^2 + 1} \to 0 \]

Thus, the range of \( \frac{3}{x^2 + 1} \) is \( (0, 3] \).

### Step 2: Identify the integer values of \( \left\lfloor \frac{3}{x^2 + 1} \right\rfloor \)

Since \( \frac{3}{x^2 + 1} \) takes values from \( 0 \) to \( 3 \), the possible integer values are \( 0, 1, 2, \) and \( 3 \).

### Step 3: Determine the intervals for each integer value

1. **For \( \left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 3 \)**:
\[ \frac{3}{x^2 + 1} \geq 3 \implies x^2 + 1 \leq 1 \implies x^2 \leq 0 \implies x = 0 \]
This holds only at \( x = 0 \).

2. **For \( \left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 2 \)**:
\[ 2 \leq \frac{3}{x^2 + 1} < 3 \implies 1 < x^2 + 1 \leq 1.5 \implies 0 < x^2 < 1 \implies 0 < x < 1 \]

3. **For \( \left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 1 \)**:
\[ 1 \leq \frac{3}{x^2 + 1} < 2 \implies 1.5 < x^2 + 1 < 3 \implies 0.5 < x^2 < 2 \implies \frac{1}{\sqrt{2}} < x < \sqrt{2} \]

4. **For \( \left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 0 \)**:
\[ \frac{3}{x^2 + 1} < 1 \implies x^2 + 1 > 3 \implies x^2 > 2 \implies x > \sqrt{2} \]

### Step 4: Set up the integral based on intervals

Now we can break the integral into parts based on the intervals we found:

\[ \int_0^\infty \left\lfloor \frac{3}{x^2 + 1} \right\rfloor dx = \int_0^1 2 \, dx + \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^\infty 0 \, dx \]

### Step 5: Calculate each integral

1. **First integral**:
\[ \int_0^1 2 \, dx = 2[x]_0^1 = 2(1 - 0) = 2 \]

2. **Second integral**:
\[ \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} 1 \, dx = [x]_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

3. **Third integral**:
\[ \int_{\sqrt{2}}^\infty 0 \, dx = 0 \]

### Step 6: Combine the results

Now, summing up the results from the integrals:
\[ \int_0^\infty \left\lfloor \frac{3}{x^2 + 1} \right\rfloor dx = 2 + \frac{1}{\sqrt{2}} + 0 = 2 + \frac{1}{\sqrt{2}} \]

### Final Result

Thus, the value of the integral is:
\[ \int_0^\infty \left\lfloor \frac{3}{x^2 + 1} \right\rfloor dx = \sqrt{2} + 1 \]

### Answer

The correct option is (B) \( \sqrt{2} + 1 \).

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