Evaluate `int _(0)^(infty) (tan^(-1)ax-tan^(-1)x)/(x)dx`, where a is a parameter.

Answer

To evaluate the integral

\[ I = \int_{0}^{\infty} \frac{\tan^{-1}(ax) - \tan^{-1}(x)}{x} \, dx, \]

where \( a \) is a parameter, we will follow these steps:

### Step 1: Define the Integral
Let

\[ I = \int_{0}^{\infty} \frac{\tan^{-1}(ax) - \tan^{-1}(x)}{x} \, dx. \]

### Step 2: Analyze the Case When \( a = 1 \)
If \( a = 1 \), then

\[ I = \int_{0}^{\infty} \frac{\tan^{-1}(x) - \tan^{-1}(x)}{x} \, dx = \int_{0}^{\infty} 0 \, dx = 0. \]

Thus, we have \( I(1) = 0 \).

### Step 3: Differentiate with Respect to \( a \)
Now, we differentiate \( I \) with respect to \( a \):

\[ \frac{dI}{da} = \int_{0}^{\infty} \frac{\partial}{\partial a} \left( \frac{\tan^{-1}(ax) - \tan^{-1}(x)}{x} \right) \, dx. \]

Using the chain rule, we find:

\[ \frac{\partial}{\partial a} \tan^{-1}(ax) = \frac{x}{1 + (ax)^2}. \]

Thus,

\[ \frac{dI}{da} = \int_{0}^{\infty} \frac{x}{1 + (ax)^2} \cdot \frac{1}{x} \, dx = \int_{0}^{\infty} \frac{1}{1 + (ax)^2} \, dx. \]

### Step 4: Simplify the Integral
Now we can simplify the integral:

\[ \frac{dI}{da} = \int_{0}^{\infty} \frac{1}{1 + (ax)^2} \, dx. \]

Using the substitution \( u = ax \) (thus \( du = a \, dx \) or \( dx = \frac{du}{a} \)), we have:

\[ \frac{dI}{da} = \int_{0}^{\infty} \frac{1}{1 + u^2} \cdot \frac{du}{a} = \frac{1}{a} \cdot \frac{\pi}{2} = \frac{\pi}{2a}. \]

### Step 5: Integrate with Respect to \( a \)
Now, we integrate \( \frac{dI}{da} \):

\[ dI = \frac{\pi}{2a} \, da. \]

Integrating both sides gives:

\[ I = \frac{\pi}{2} \ln a + C, \]

where \( C \) is a constant of integration.

### Step 6: Determine the Constant \( C \)
We know that \( I(1) = 0 \):

\[ 0 = \frac{\pi}{2} \ln(1) + C \implies C = 0. \]

### Final Result
Thus, we have:

\[ I = \frac{\pi}{2} \ln a. \]

Therefore, the final answer is:

\[ \int_{0}^{\infty} \frac{\tan^{-1}(ax) - \tan^{-1}(x)}{x} \, dx = \frac{\pi}{2} \ln a. \]

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