Question

$\displaystyle \sf \int_{1}^{ \infty} \frac{1}{ {x}^{p} } \: dx$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \tt \int^{ \infty }_{1} \frac{1}{ {x}^{p} } \: dx$

$\displaystyle = \tt \int^{ \infty }_{1} {x}^{ - p} \: dx$

$\displaystyle = \tt \ \left[ \dfrac{ {x}^{ - p + 1}}{ - p + 1} \right]^{ \infty }_{1}$

$\displaystyle = \tt \ \left[ \dfrac{ 1}{ (1- p) \cdot \: {x}^{1 - p} } \right]^{ \infty }_{1}$

$\displaystyle = \tt \ \dfrac{ 1}{ (1- p) \cdot \: \infty } - \dfrac{1}{1 - p}$

$\displaystyle = \tt \dfrac{1}{p - 1 }$

Answer 2

Answer:

We have,

$\red \displaystyle \tt \int^{ \infty }_{1} \frac{1}{ {x}^{p} } \: dx$

$\blue \displaystyle = \tt \int^{ \infty }_{1} {x}^{ - p} \: dx$

$\green \displaystyle = \tt \ \left[ \dfrac{ {x}^{ - p + 1}}{ - p + 1} \right]^{ \infty }_{1}$

$\displaystyle = \tt \ \left[ \dfrac{ 1}{ (1- p) \cdot \: {x}^{1 - p} } \right]^{ \infty }_{1}$

$\displaystyle = \tt \ \dfrac{ 1}{ (1- p) \cdot \: \infty } - \dfrac{1}{1 - p}$

$\displaystyle = \tt \dfrac{1}{p - 1 }$

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