What is the integral from 0 to a of (m/(2pi(a-x)))^5 ?

The integral from 0 to a of (m/(2pi(a-x)))^5 is diverges

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integral from 0 to a of (m/(2pi(a-x)))^5

\int_{0}^{a} (\frac{m}{2 π ( a - x )})^{5} d x

diverges

Solution steps

\int_{0}^{a} (\frac{m}{2 π ( a - x )})^{5} d x

Apply u-substitution

: \int_{a}^{0} - \frac{m ^{5}}{32 π ^{5} u ^{5}} d u

= \int_{a}^{0} - \frac{m ^{5}}{32 π ^{5} u ^{5}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{m ^{5}}{32 π ^{5}} \cdot \int_{a}^{0} \frac{1}{u ^{5}} d u

Apply the Power Rule

: [- \frac{1}{4 u ^{4}}]_{a}^{0}

= - \frac{m ^{5}}{32 π ^{5}} [- \frac{1}{4 u ^{4}}]_{a}^{0}

Compute the boundaries:

diverges

= diverges

y = 3 x^{4} - 3 x^{2}, y = 3 x^{2}

x \to 2 lim (\frac{2 x + 4}{x - 7})

\int_{0}^{\infty} e^{- 0.2 x} d x

\int 1 0^{- 3 x} d x

\frac{d}{d x} ((e^{x})^{3})

What is the integral from 0 to a of (m/(2pi(a-x)))^5 ?
The integral from 0 to a of (m/(2pi(a-x)))^5 is diverges