If `f(x)=int_(0)^(x) log ((1-t)/(1+t)) dt`, then discuss whether even or odd?

Answer

To determine whether the function \( f(x) = \int_{0}^{x} \log\left(\frac{1-t}{1+t}\right) dt \) is even or odd, we will follow these steps:

### Step 1: Define the function and its negative argument
We start with the function defined as:
\[ f(x) = \int_{0}^{x} \log\left(\frac{1-t}{1+t}\right) dt \]

Next, we will evaluate \( f(-x) \):
\[ f(-x) = \int_{0}^{-x} \log\left(\frac{1-t}{1+t}\right) dt \]

### Step 2: Change of variable in the integral
To evaluate \( f(-x) \), we perform a change of variable. Let \( t = -r \), then \( dt = -dr \). Changing the limits accordingly:
- When \( t = 0 \), \( r = 0 \)
- When \( t = -x \), \( r = x \)

Thus, we have:
\[ f(-x) = \int_{0}^{x} \log\left(\frac{1-(-r)}{1+(-r)}\right)(-dr) = -\int_{0}^{x} \log\left(\frac{1+r}{1-r}\right) dr \]

### Step 3: Simplifying the logarithmic expression
Using the property of logarithms, we can rewrite:
\[ \log\left(\frac{1+r}{1-r}\right) = -\log\left(\frac{1-t}{1+t}\right) \]
Thus, we have:
\[ f(-x) = -\left(-\int_{0}^{x} \log\left(\frac{1-t}{1+t}\right) dt\right) = \int_{0}^{x} \log\left(\frac{1-t}{1+t}\right) dt = f(x) \]

### Step 4: Conclusion
Since \( f(-x) = f(x) \), we conclude that the function \( f(x) \) is an **even function**.

### Summary
The function \( f(x) = \int_{0}^{x} \log\left(\frac{1-t}{1+t}\right) dt \) is even because \( f(-x) = f(x) \).

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