Question

QUESTION 6 Sanna and Obadia argued about the definition of $\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$ Sanna split the integral as $\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx = \int_{-\infty}^{0} \frac{e^{-x}}{x^2 - 1} dx + \int_{0}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$ Obadia split it differently as $\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx = \int_{-\infty}^{-1} \frac{e^{-x}}{x^2 - 1} dx + \int_{-1}^{1} \frac{e^{-x}}{x^2 - 1} dx + \int_{1}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$ Which of the two would you agree with? Clearly explain your choice.

          QUESTION 6
Sanna and Obadia argued about the definition of
$\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$
Sanna split the integral as
$\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx = \int_{-\infty}^{0} \frac{e^{-x}}{x^2 - 1} dx + \int_{0}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$
Obadia split it differently as
$\int_{-\infty}^{\infty} \frac{e^{-x}}{x^2 - 1} dx = \int_{-\infty}^{-1} \frac{e^{-x}}{x^2 - 1} dx + \int_{-1}^{1} \frac{e^{-x}}{x^2 - 1} dx + \int_{1}^{\infty} \frac{e^{-x}}{x^2 - 1} dx$
Which of the two would you agree with? Clearly explain your choice.

QUESTION 6
Sanna and Obadia argued about the definition of
∫-∞^∞(e^-x)/(x^2 - 1) dx
Sanna split the integral as
∫-∞^∞(e^-x)/(x^2 - 1) dx = ∫-∞^0(e^-x)/(x^2 - 1) dx + ∫0^∞(e^-x)/(x^2 - 1) dx
Obadia split it differently as
∫-∞^∞(e^-x)/(x^2 - 1) dx = ∫-∞^-1(e^-x)/(x^2 - 1) dx + ∫-1^1(e^-x)/(x^2 - 1) dx + ∫1^∞(e^-x)/(x^2 - 1) dx
Which of the two would you agree with? Clearly explain your choice.

Added by Lauren A.

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