Let $$g(t) = \int_{-\pi/2}^{\pi/2} (\cos \frac{\pi}{4}t + f(x))dx$$, where $$f(x) = \log_e(x + \sqrt{x^2+1})$$, $$x \in R$$. Then which one of the following is correct?

We have $$g(t) = \int_{-\pi/2}^{\pi/2} \left(\cos\frac{\pi}{4}t + f(x)\right) dx$$, where $$f(x) = \log_e\!\left(x + \sqrt{x^2+1}\right)$$.

First, we note that $$f(x)$$ is an odd function: $$f(-x) = \log_e\!\left(-x + \sqrt{x^2+1}\right)$$, and since $$(-x + \sqrt{x^2+1})(x + \sqrt{x^2+1}) = x^2+1-x^2 = 1$$, we get $$f(-x) = -\log_e(x+\sqrt{x^2+1}) = -f(x)$$. Therefore, $$\int_{-\pi/2}^{\pi/2} f(x)\,dx = 0.$$

Also, $$\cos\frac{\pi}{4}t$$ is a constant with respect to $$x$$, so: $$g(t) = \cos\frac{\pi t}{4} \cdot \int_{-\pi/2}^{\pi/2} dx + 0 = \pi\cos\frac{\pi t}{4}.$$

Evaluating: $$g(0) = \pi\cos 0 = \pi$$ and $$g(1) = \pi\cos\frac{\pi}{4} = \frac{\pi}{\sqrt{2}}.$$

Therefore, $$\sqrt{2}\,g(1) = \sqrt{2} \cdot \frac{\pi}{\sqrt{2}} = \pi = g(0)$$, i.e., $$\sqrt{2}\,g(1) = g(0)$$.