For what values of p the integral ∫_0 ^1 x^p ln x dx converge?


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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
	$lnx×(x^(p+1) /(p+1))−∫(1/x)×(x^(p+1) /(p+1))dx ∣((x^(p+1) lnx)/(p+1))−(1/(p+1))×(x^(p+1) /(p+1))∣_0 ^1 =((1×0)/(p+1))−(1/((p+1)^2 )) =((−1)/((p+1)^2 )) p≠−1 so i think value of p is R−{−1} pls check$
	$l n x \times \frac{x^{p + 1}}{p + 1} - \int \frac{1}{x} \times \frac{x^{p + 1}}{p + 1} d x$ $∣ \frac{x^{p + 1} l n x}{p + 1} - \frac{1}{p + 1} \times \frac{x^{p + 1}}{p + 1} ∣_{0}^{1}$ $= \frac{1 \times 0}{p + 1} - \frac{1}{{(p + 1)}^{2}}$ $= \frac{- 1}{{(p + 1)}^{2}} p \neq - 1$ $s o i t h i n k v a l u e o f p i s R - {- 1} p l s c h e c k$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

	$t h a n k y o u s i r . . . i s m y a n s w e r c o r r e c t . . .$
	Commented by Joel578 last updated on 05/Mar/19
	$Sir I think p should be ≤ −1 I =lim_(t→0^+ ) [((x^(p+1) (ln x))/(p + 1)) − (x^(p+1) /((p + 1)^2 ))]_t ^1 ⇒ lim_(t→0^+ ) {(0 − (1/((p + 1)^2 ))) − (((t^(p+1) (ln t))/(p + 1)) − (t^(p+1) /((p + 1)^2 )))} ⇒ − (1/((p + 1)^2 )) − lim_(t→0^+ ) {(t^(p+1) /(p + 1))(ln t − (1/(p + 1)))} when p + 1 = 0 ⇔ p = −1, I diverge when p + 1 < 0 ⇔ p < − 1, (p + 1) is negative, which mean t^(p+1) → ∞ as t → 0^+ . Hence I = −(1/((p + 1)^2 )) − ∞ = −∞ (diverge)$
	$Sir I think p should be ⩽ - 1$ $I = \lim_{t \to 0^{+}} {[\frac{x^{p + 1} (\ln x)}{p + 1} - \frac{x^{p + 1}}{{(p + 1)}^{2}}]}_{t}^{1}$ $\Rightarrow \lim_{t \to 0^{+}} {(0 - \frac{1}{{(p + 1)}^{2}}) - (\frac{t^{p + 1} (\ln t)}{p + 1} - \frac{t^{p + 1}}{{(p + 1)}^{2}})}$ $\Rightarrow - \frac{1}{{(p + 1)}^{2}} - \lim_{t \to 0^{+}} {\frac{t^{p + 1}}{p + 1} (\ln t - \frac{1}{p + 1})}$ $when p + 1 = 0 \Leftrightarrow p = - 1, I diverge$ $when p + 1 < 0 \Leftrightarrow p < - 1, (p + 1) is negative,$ $which mean t^{p + 1} \to \infty as t \to 0^{+} .$ $Hence I = - \frac{1}{{(p + 1)}^{2}} - \infty = - \infty (diverge)$
	Commented by Joel578 last updated on 05/Mar/19

	$please correct if I am wrong$
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