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Question Number 217691 by PaulDirac last updated on 18/Mar/25

	Answered by mr W last updated on 19/Mar/25
	$let a=9^9^9 I=∫_0 ^π ln (ax)dx =(1/a)∫_0 ^π ln (ax)d(ax) =(1/a)∫_0 ^(aπ) ln t dt =(1/a){[t ln t]_0 ^(aπ) −∫_0 ^(aπ) dt} =(1/a)[t(ln t−1)]_0 ^(aπ) =(1/a)×aπ(ln aπ−1)−(1/a)lim_(t→0) t(ln t−1) =π(ln aπ−1)−(1/a)lim_(t→0) (ln t^t −t) ^(∗)) =π(ln aπ−1) =π[ln (9^9^9 π)−1] ✓ ^(∗)) Note: lim_(x→0) x^x =1 ⇒lim_(x→0) (ln x^x )=0$
	$l e t a = 9^{9^{9}}$ $I = \int_{0}^{π} \ln (a x) d x$ $= \frac{1}{a} \int_{0}^{π} \ln (a x) d (a x)$ $= \frac{1}{a} \int_{0}^{a π} \ln t d t$ $= \frac{1}{a} {{[t \ln t]}_{0}^{a π} - \int_{0}^{a π} d t}$ $= \frac{1}{a} {[t (\ln t - 1)]}_{0}^{a π}$ $= \frac{1}{a} \times a π (\ln a π - 1) - \frac{1}{a} \lim_{t \to 0} t (\ln t - 1)$ $= π (\ln a π - 1) - \frac{1}{a} \lim_{t \to 0} (\ln t^{t} - t)^{)}$ $= π (\ln a π - 1)$ $= π [\ln (9^{9^{9}} π) - 1] ✓$ $^{)} N o t e :$ $\lim_{x \to 0} x^{x} = 1 \Rightarrow \lim_{x \to 0} (\ln x^{x}) = 0$
	Commented by SdC355 last updated on 19/Mar/25

	$sorry, i was careless .$
	Commented by mr W last updated on 19/Mar/25

	$t h e q u e s t i o n i s c l e a r l y$ $\log_{e} 9^{9^{9}} x = \log_{e} (9^{9^{9}} x) = \ln (9^{9^{9}} x) .$ $n o t h i n g e l s e i s m e a n t .$
	Commented by mr W last updated on 19/Mar/25

	${i t}^{'} s a l r i g h t! n o n e e d f o r b e i n g s o r r y!$
	Commented by SdC355 last updated on 19/Mar/25

	$? ? ?$ $Prime causes double exponent: use braces to clarify$ $\ln (e^{9^{9^{9}}} z) ? ?$
	Answered by Wuji last updated on 19/Mar/25

	$\underset{0}{\int^{π}} {l o g}_{e} (9^{9^{9}}) x d x$ ${l o g}_{e} (9^{9^{9}}) \underset{0}{\int^{π}} x d x = {l o g}_{e} (9^{9^{9}}) \frac{x^{2}}{2} ∣_{0}^{π}$ $= {l o g}_{e} (9^{9^{9}}) \frac{π^{2}}{2}$ $= 9^{9} {l o g}_{e} (9) \frac{π^{2}}{2}$
	Commented by mr W last updated on 20/Mar/25

	$i t h i n k n o b o d y w i l l r e a l l y m e a n$ $w i t h \int \sin 2 x d x a s$ $\int (\sin 2) x d x = (\sin 2) \int x d x$
	Commented by Wuji last updated on 22/Mar/25

	$m y a p o l o g i e s, s i r . i t h o u g h t t h e v a r i a b l e$ $i s n o t a t t a c h e d w i t h t h e c o n s t a n t .$
	Answered by Wuji last updated on 22/Mar/25
	$Thanks to Mr. W for pointing out my errors ∫_0 ^π log_e (9^9^9 x)dx =∫_0 ^π ln(9^9^9 x)dx I(a)=∫_0 ^π ln(ax)dx {a=9^9^9 } (d/da)I(a)=∫_0 ^π (d/da)ln(ax)dx=∫_0 ^π (1/(ax))xdx=∫_0 ^π (1/a)dx=(π/a) I(a)=∫_0 ^π (π/a)da =πln(a)+C I(1)=∫_0 ^π ln(x)dx =xln(x)−x+C I(1)=[πln(π)−π]−lim_(x→0^+ ) (xln(x)−x)=πln(π)−π I(a)=πln(a)+πln(π)−π=π(ln(aπ)−1) a=9^9^9 I=π[ln(9^9^9 π)−1] =π(9^9 ln(9π)−1)$
	$T h a n k s t o M r . W f o r p o i n t i n g o u t m y e r r o r s$ $\int_{0}^{π} {l o g}_{e} (9^{9^{9}} x) d x = \int_{0}^{π} l n (9^{9^{9}} x) d x$ $I (a) = \int_{0}^{π} l n (a x) d x {a = 9^{9^{9}}}$ $\frac{d}{d a} I (a) = \int_{0}^{π} \frac{d}{d a} l n (a x) d x = \int_{0}^{π} \frac{1}{a x} x d x = \int_{0}^{π} \frac{1}{a} d x = \frac{π}{a}$ $I (a) = \int_{0}^{π} \frac{π}{a} d a = π l n (a) + C$ $I (1) = \int_{0}^{π} l n (x) d x = x l n (x) - x + C$ $I (1) = [π l n (π) - π] - \lim_{x \to 0^{+}} (x l n (x) - x) = π l n (π) - π$ $I (a) = π l n (a) + π l n (π) - π = π (l n (a π) - 1)$ $a = 9^{9^{9}}$ $I = π [l n (9^{9^{9}} π) - 1] = π (9^{9} l n (9 π) - 1)$
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