Question-217691

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

Answered by mr W last updated on 19/Mar/25

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

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