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Just one of these four integrals converges. Which?

  • A.\( \displaystyle\int_{1}^{\infty} \frac{dx}{\sqrt{x}} \)
  • B.\( \displaystyle\int_{1}^{\infty} \frac{dx}{x^{3}} \)
  • C.\( \displaystyle\int_{1}^{\infty} \frac{dx}{x} \)
  • D.\( \displaystyle\int_{0}^{1} \frac{dx}{x} \)

Answer: B

For the tail at infinity, convergence needs the exponent above one, so only \( x^{-3} \) qualifies; \( x^{-1} \) and \( x^{-1/2} \) both diverge. The integral of \( 1/x \) from zero to one diverges as well, since \( \ln x \to -\infty \) at the lower end, so it fails at the other kind of singularity

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