Question

\(\displaystyle \int_{0}^{\infty}\left(x^{12}+x^{-12}\right)\frac{\log x}{x}\,dx=\)

• A. \(0\)
• B. \(1\)
• C. \(\log 2\)
• D. \(e^2\)

Hint:

For integrals over \((0,\infty)\), the substitution \[ x=\frac{1}{t} \] is very useful. If the transformed integral becomes the negative of the original integral, then the value is \(0\).

Answer 1

The Correct Option is A

Step 1: Observe the given integral.
Let \[ I=\int_{0}^{\infty}\left(x^{12}+x^{-12}\right)\frac{\log x}{x}\,dx \]

Step 2: Use the substitution \(x=\dfrac{1}{t}\).
Put \[ x=\frac{1}{t} \] Then, \[ dx=-\frac{1}{t^2}\,dt \] Also, \[ \log x=\log\frac{1}{t}=-\log t \] and \[ x^{12}+x^{-12}=t^{-12}+t^{12} \] \[ \frac{1}{x}=t \]

Step 3: Transform the integral.
Using the substitution, the limits change as \[ x=0 \Rightarrow t=\infty \] and \[ x=\infty \Rightarrow t=0 \] Therefore, \[ I=\int_{\infty}^{0}(t^{-12}+t^{12})(-\log t)t\left(-\frac{1}{t^2}\right)\,dt \] \[ I=-\int_{0}^{\infty}(t^{12}+t^{-12})\frac{\log t}{t}\,dt \] Thus, \[ I=-I \]

Step 4: Solve for \(I\).
\[ I=-I \] \[ 2I=0 \] \[ I=0 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]