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Вычислить несобственный интеграл или установить его расходимость int_(0)^(1)cos ((pi)/(1-x))*(dx)/((1-x)^(2))

Вычислить несобственный интеграл или установить его расходимость\newline∫01cos⁡π1−x⋅dx(1−x)2 \int_{0}^{1} \cos \frac{\pi}{1-x} \cdot \frac{d x}{(1-x)^{2}}

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Q. Вычислить несобственный интеграл или установить его расходимость\newline∫01cos⁡π1−x⋅dx(1−x)2 \int_{0}^{1} \cos \frac{\pi}{1-x} \cdot \frac{d x}{(1-x)^{2}}
  1. Make Substitution: Let's make a substitution to simplify the integral.\newlineLet u=1−xu = 1 - x, then du=−dxdu = -dx.\newlineRewrite the integral in terms of uu.\newline∫01cos⁡(π1−x)dx(1−x)2=∫10cos⁡(πu)−duu2\int_{0}^{1}\cos\left(\frac{\pi}{1-x}\right)\frac{dx}{(1-x)^{2}} = \int_{1}^{0}\cos\left(\frac{\pi}{u}\right)\frac{-du}{u^{2}}
  2. Rewrite Integral: Since we have a negative sign from dudu and the limits are swapped, we can remove the negative sign by swapping the limits back.\newline∫10cos⁡(π/u)−duu2=∫01cos⁡(π/u)duu2\int_{1}^{0}\cos(\pi/u)\frac{-du}{u^2} = \int_{0}^{1}\cos(\pi/u)\frac{du}{u^2}
  3. Evaluate Integral: Now, let's evaluate the integral. \newline∫01cos⁡(πu)duu2\int_{0}^{1}\cos(\frac{\pi}{u})\frac{du}{u^2}\newlineThis is an improper integral because the integrand becomes unbounded as uu approaches 00.\newlineWe need to take the limit as uu approaches 00 from the right.\newlinelim⁡t→0+∫t1cos⁡(πu)duu2\lim_{t\to0^+} \int_{t}^{1}\cos(\frac{\pi}{u})\frac{du}{u^2}
  4. Find Antiderivative: To evaluate the limit, we need to find the antiderivative of cos⁡(π/u)/u2\cos(\pi/u)/u^2. However, this integral does not have an elementary antiderivative, so we cannot proceed with standard integration techniques.
  5. Determine Convergence: Since we cannot find an antiderivative, we need to determine if the integral converges or diverges by other means.\newlineWe can compare it to a known divergent integral.\newlineAs uu approaches 00, 1u2\frac{1}{u^2} approaches infinity faster than cos⁡(π/u)\cos(\pi/u), which is bounded between −1-1 and 11.\newlineTherefore, the integral behaves like 1u2\frac{1}{u^2} near 00, which is a known divergent pp-integral for p≤1p \leq 1.
  6. Conclude Divergence: Since the integral behaves like a divergent pp-integral near 00, the original integral also diverges.\newlineTherefore, we can conclude that the improper integral diverges.

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