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

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

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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