If `f(x)={xcosx+(log)_e((1-x)/(1+x))a; x=0; x!=0` is odd, then `a_______,`

Explanation

To determine the value of \( a \) such that the function

\[ f(x) = \begin{cases} x \cos x + a \log_e \left( \frac{1-x}{1+x} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

is an odd function, we need to use the property of odd functions. A function \( f(x) \) is odd if

\[ f(-x) = -f(x) \]

for all \( x \).

### Step 1: Calculate \( f(-x) \)

For \( x \neq 0 \):

\[ f(-x) = -x \cos(-x) + a \log_e \left( \frac{1+x}{1-x} \right) \]

Since \( \cos(-x) = \cos(x) \), we have:

\[ f(-x) = -x \cos x + a \log_e \left( \frac{1+x}{1-x} \right) \]

### Step 2: Simplify the logarithmic term

Using the property of logarithms, we can rewrite the logarithmic term:

\[ \log_e \left( \frac{1+x}{1-x} \right) = -\log_e \left( \frac{1-x}{1+x} \right) \]

Thus,

\[ f(-x) = -x \cos x - a \log_e \left( \frac{1-x}{1+x} \right) \]

### Step 3: Set up the equation for odd function property

Now, we need to set \( f(-x) = -f(x) \):

\[ -x \cos x - a \log_e \left( \frac{1-x}{1+x} \right) = -\left( x \cos x + a \log_e \left( \frac{1-x}{1+x} \right) \right) \]

### Step 4: Simplify the equation

This simplifies to:

\[ -x \cos x - a \log_e \left( \frac{1-x}{1+x} \right) = -x \cos x - a \log_e \left( \frac{1-x}{1+x} \right) \]

This means:

\[ -a \log_e \left( \frac{1-x}{1+x} \right) = -a \log_e \left( \frac{1-x}{1+x} \right) \]

### Step 5: Solve for \( a \)

For the equation to hold true for all \( x \), we must have:

\[ a = -a \]

This implies:

\[ 2a = 0 \implies a = 0 \]

### Conclusion

Thus, the value of \( a \) is

\[ \boxed{0} \]