\( \int _ { 0 } ^ { + \infty } ( e ^ { \frac { x } { 1 + x } } - \cos x ) ^ { - \frac { 1 } { 2 } } \tan ( \frac { 1 } { x + 1 } ) d x \)

To solve the integral \( I = \int _ { 0 } ^ { + \infty } ( e ^ { \frac { x } { 1 + x } } - \cos x ) ^ { - \frac { 1 } { 2 } } \tan ( \frac { 1 } { x + 1 } ) d x \), we can begin by analyzing the behavior of the integrand as \( x \) approaches 0 and as \( x \) approaches infinity. 1. **Behavior as \( x \to 0 \)**: - We have \( e^{\frac{x}{1+x}} \) approximating \( e^0 = 1 \), hence \( e^{\frac{x}{1+x}} - \cos x \approx 1 - (1 - \frac{x^2}{2}) = \frac{x^2}{2} \). - Therefore, as \( x \to 0 \): \[ (e^{\frac{x}{1+x}} - \cos x)^{-1/2} \approx \left(\frac{x^2}{2}\right)^{-1/2} = \sqrt{\frac{2}{x^2}} = \frac{\sqrt{2}}{x} \] - Also, \( \tan\left(\frac{1}{x + 1}\right) \approx \frac{1}{x + 1} \approx 1 \) as \( x \to 0 \). - Thus, the integrand behaves like: \[ \frac{\sqrt{2}}{x} \cdot 1 = \frac{\sqrt{2}}{x} \] - This indicates a singularity at \( x = 0 \) but the integral: \[ \int_0^c \frac{\sqrt{2}}{x} \, dx \quad \text{ diverges as } c \to 0. \] 2. **Behavior as \( x \to \infty \)**: - Here, \( e^{\frac{x}{1+x}} \) approaches \( e^1 = e \), and \( \cos x \) oscillates between -1 and 1, so: \[ e^{\frac{x}{1+x}} - \cos x \to e + 1 \text{ (which is a positive constant)}. \] - Consequently, \( (e^{\frac{x}{1+x}} - \cos x)^{-1/2} \) approaches a constant: \[ (e + 1)^{-1/2}. \] - The \( \tan\left(\frac{1}{x + 1}\right) \) tends toward 0, since as \( x \to \infty \), \( \frac{1}{x + 1} \to 0 \). - Thus: \[ \tan\left(\frac{1}{x+1}\right) \sim \frac{1}{x + 1} \to 0. \] - The integrand approaches 0 as \( x \to \infty \). Ultimately, the contribution near zero leads us to conclude that the integral diverges due to the behavior at the lower limit. Thus, we find: \[ I \text{ diverges}. \]