The value of the integral $$\displaystyle\int_0^{\infty} \frac{\log_e (x)}{x^2 + 4} \, dx$$ is:

To solve the integral $$I = \int_{0}^{\infty} \frac{\log_e(x)}{x^2 + 4} dx$$, we can use a clever substitution method that leverages the properties of logarithms and symmetry.

Let's use the substitution $$x = \frac{4}{t}$$.

Differentiating both sides gives:

$$dx = -\frac{4}{t^2} dt$$

Changing the limits:

• As $$x \to 0$$, $$t \to \infty$$
• As $$x \to \infty$$, $$t \to 0$$

Substituting these values into the original integral:

$$I = \int_{\infty}^{0} \frac{\ln(4/t)}{(4/t)^2 + 4} \left( -\frac{4}{t^2} \right) dt$$

Using the negative sign to flip the limits back to $$(0, \infty)$$:

$$I = \int_{0}^{\infty} \frac{\ln(4) - \ln(t)}{\frac{16}{t^2} + 4} \cdot \frac{4}{t^2} dt$$

$$I = \int_{0}^{\infty} \frac{\ln(4) - \ln(t)}{\frac{16 + 4t^2}{t^2}} \cdot \frac{4}{t^2} dt$$

The $$t^2$$ terms cancel out:

$$I = \int_{0}^{\infty} \frac{4(\ln 4 - \ln t)}{4(4 + t^2)} dt$$

$$I = \int_{0}^{\infty} \frac{\ln 4}{t^2 + 4} dt - \int_{0}^{\infty} \frac{\ln t}{t^2 + 4} dt$$

Notice that the second part of the expression, $$\int_{0}^{\infty} \frac{\ln t}{t^2 + 4} dt$$, is exactly the same as our original integral $$I$$ (just using a different dummy variable $$t$$).

So, we have:

$$I = \int_{0}^{\infty} \frac{\ln 4}{t^2 + 4} dt - I$$

$$2I = \ln(4) \int_{0}^{\infty} \frac{1}{t^2 + 4} dt$$

The integral $$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$$. Here, $$a=2$$:

$$2I = \ln(4) \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_{0}^{\infty}$$

$$2I = \ln(2^2) \cdot \frac{1}{2} \left[ \tan^{-1}(\infty) - \tan^{-1}(0) \right]$$

$$2I = 2\ln(2) \cdot \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right]$$

$$2I = \ln(2) \cdot \frac{\pi}{2}$$

$$I = \frac{\pi \ln(2)}{4}$$

Correct Option:

The value of the integral is B: $$\frac{\pi \log_e(2)}{4}$$.