Question

$\int_{0}^{\infty} \frac{\ln x}{x^2 + kx + k^2} dx$, where k > 0.

Answer 1

$\frac{2\pi \ln k}{3\sqrt{3}k}$

Answer 2

The problem asks to evaluate the definite integral $I = \int_{0}^{\infty} \frac{\ln x}{x^2 + kx + k^2} dx$, where $k > 0$.

Explanation of the Solution:

1. Identify the general form and apply a known property: The integral is of the form $\int_{0}^{\infty} \frac{\ln x}{x^2+ax+b} dx$. A common property for such integrals states that if $b>0$, then $I = \frac{\ln b}{2} \int_{0}^{\infty} \frac{1}{x^2+ax+b} dx$. In our given integral, $a=k$ and $b=k^2$. Since $k>0$, $b=k^2>0$. Applying the property: $I = \frac{\ln(k^2)}{2} \int_{0}^{\infty} \frac{1}{x^2 + kx + k^2} dx$ $I = \frac{2 \ln k}{2} \int_{0}^{\infty} \frac{1}{x^2 + kx + k^2} dx$ $I = \ln k \int_{0}^{\infty} \frac{1}{x^2 + kx + k^2} dx$

2. Evaluate the remaining integral: We need to evaluate $J = \int_{0}^{\infty} \frac{1}{x^2 + kx + k^2} dx$. Complete the square in the denominator: $x^2 + kx + k^2 = x^2 + kx + \left(\frac{k}{2}\right)^2 - \left(\frac{k}{2}\right)^2 + k^2$ $= \left(x + \frac{k}{2}\right)^2 + k^2 - \frac{k^2}{4}$ $= \left(x + \frac{k}{2}\right)^2 + \frac{3k^2}{4}$ $= \left(x + \frac{k}{2}\right)^2 + \left(\frac{\sqrt{3}k}{2}\right)^2$

   Now, substitute $u = x + \frac{k}{2}$. Then $du = dx$. The limits of integration change: When $x=0$, $u = 0 + \frac{k}{2} = \frac{k}{2}$. When $x=\infty$, $u = \infty + \frac{k}{2} = \infty$. So, the integral becomes: $J = \int_{k/2}^{\infty} \frac{1}{u^2 + \left(\frac{\sqrt{3}k}{2}\right)^2} du$

   This is a standard integral of the form $\int \frac{1}{y^2+A^2} dy = \frac{1}{A} \arctan\left(\frac{y}{A}\right)$. Here, $A = \frac{\sqrt{3}k}{2}$. $J = \left[ \frac{1}{\frac{\sqrt{3}k}{2}} \arctan\left(\frac{u}{\frac{\sqrt{3}k}{2}}\right) \right]_{k/2}^{\infty}$ $J = \frac{2}{\sqrt{3}k} \left[ \arctan\left(\frac{2u}{\sqrt{3}k}\right) \right]_{k/2}^{\infty}$ $J = \frac{2}{\sqrt{3}k} \left( \lim_{u \to \infty} \arctan\left(\frac{2u}{\sqrt{3}k}\right) - \arctan\left(\frac{2(k/2)}{\sqrt{3}k}\right) \right)$ $J = \frac{2}{\sqrt{3}k} \left( \arctan(\infty) - \arctan\left(\frac{k}{\sqrt{3}k}\right) \right)$ $J = \frac{2}{\sqrt{3}k} \left( \frac{\pi}{2} - \arctan\left(\frac{1}{\sqrt{3}}\right) \right)$ $J = \frac{2}{\sqrt{3}k} \left( \frac{\pi}{2} - \frac{\pi}{6} \right)$ $J = \frac{2}{\sqrt{3}k} \left( \frac{3\pi - \pi}{6} \right)$ $J = \frac{2}{\sqrt{3}k} \left( \frac{2\pi}{6} \right)$ $J = \frac{2}{\sqrt{3}k} \left( \frac{\pi}{3} \right)$ $J = \frac{2\pi}{3\sqrt{3}k}$

3. Combine the results: Substitute the value of $J$ back into the expression for $I$: $I = \ln k \cdot J$ $I = \ln k \cdot \frac{2\pi}{3\sqrt{3}k}$ $I = \frac{2\pi \ln k}{3\sqrt{3}k}$