For-what-values-of-p-the-integral-0-1-x-p-ln-x-dx-converge-

Question Number 55856 by Joel578 last updated on 05/Mar/19

For what values of p the integral

\int_{0}^{1} x^{p} \ln x d x

converge ?

Commented by maxmathsup by imad last updated on 05/Mar/19

l e t A_{p} = \int_{0}^{1} x^{p} l n (x) d x c h a n g e m e n t l n (x) = - t g i v e x = e^{- t}

A_{p} = - \int_{0}^{\infty} {(e^{- t})}^{p} (t) e^{- t} d t = - \int_{0}^{\infty} t e^{- (p + 1) t} d t

i f p > - 1 A_{p} c o n v e r g e s i f p ⩽ - 1 {l i m}_{t \to + \infty} t e^{- (p + 1) t} = + \infty \Rightarrow A_{p} d i v e r g e s

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

l n x \times \frac{x^{p + 1}}{p + 1} - \int \frac{1}{x} \times \frac{x^{p + 1}}{p + 1} d x

∣ \frac{x^{p + 1} l n x}{p + 1} - \frac{1}{p + 1} \times \frac{x^{p + 1}}{p + 1} ∣_{0}^{1}

= \frac{1 \times 0}{p + 1} - \frac{1}{{(p + 1)}^{2}}

= \frac{- 1}{{(p + 1)}^{2}} p \neq - 1

s o i t h i n k v a l u e o f p i s R - {- 1} p l s c h e c k

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

t h a n k y o u s i r \dots i s m y a n s w e r c o r r e c t \dots

Commented by Joel578 last updated on 05/Mar/19

Sir I think p should be ⩽ - 1

I = \lim_{t \to 0^{+}} {[\frac{x^{p + 1} (\ln x)}{p + 1} - \frac{x^{p + 1}}{{(p + 1)}^{2}}]}_{t}^{1}

\Rightarrow \lim_{t \to 0^{+}} {(0 - \frac{1}{{(p + 1)}^{2}}) - (\frac{t^{p + 1} (\ln t)}{p + 1} - \frac{t^{p + 1}}{{(p + 1)}^{2}})}

\Rightarrow - \frac{1}{{(p + 1)}^{2}} - \lim_{t \to 0^{+}} {\frac{t^{p + 1}}{p + 1} (\ln t - \frac{1}{p + 1})}

when p + 1 = 0 \Leftrightarrow p = - 1, I diverge

when p + 1 < 0 \Leftrightarrow p < - 1, (p + 1) is negative,

which mean t^{p + 1} \to \infty as t \to 0^{+} .

Hence I = - \frac{1}{{(p + 1)}^{2}} - \infty = - \infty (diverge)

Commented by Joel578 last updated on 05/Mar/19

please correct if I am wrong

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