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Question Number 2088 by Yozzi last updated on 01/Nov/15

S h o w t h a t \forall n \in Z^{+}

\int_{0}^{π / 4} {t a n}^{2 n + 1} θ d θ = {(- 1)}^{n} (\frac{1}{2} l n 2 + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{2 m}) .

D e d u c e t h a t \underset{1}{\sum^{\infty}} \frac{{(- 1)}^{m - 1}}{2 m} = 0.5 l n 2 .

S h o w a l s o t h a t \underset{1}{\sum^{\infty}} \frac{{(- 1)}^{m - 1}}{2 m - 1} = \frac{π}{4} .

Commented by Yozzi last updated on 02/Nov/15

{I t}^{'} s a n o l d (1990) a d m i s s i o n e x a m

q u e s t i o n a n d I {w a s n}^{'} t s u r e w h e r e

t h e {(- 1)}^{n} f a c t o r c a m e f r o m .

B e f o r e t h e p a r t I g a v e h e r e t h e

q u e s t i o n a s k e d a b o u t t h e s k e t c h e s

o f t h e g r a p h o f y = {t a n}^{k} θ f o r

k = 1 a n d k ⋙ 1 w h e r e x \in [0, π / 4] .

Y o u r m e t h o d a p p e a r s t o b e c o r r e c t .

I^{'} m w o n d e r i n g w h a t w e n t w r o n g \dots

Commented by prakash jain last updated on 03/Nov/15

\tan^{2 n + 1} θ = {(\tan^{2} θ)}^{n} \tan θ

= {(\sec^{2} θ - 1)}^{n} \tan θ

= {(- 1)}^{n} {(1 - \sec^{2} θ)}^{n} \tan θ

= {(- 1)}^{n} [\tan θ + \underset{m = 1}{\sum^{n}}^{n} C_{m} {(- 1)}^{m} \sec^{2 m} θ \tan θ]

= {(- 1)}^{n} [\tan θ + \underset{m = 1}{\sum^{n}} {(- 1)}^{m} \sec^{2 m - 1} θ \cdot \sec θ \cdot \tan θ]

Integrating

= {(- 1)}^{n} [\ln \sec θ + \underset{m = 1}{\sum^{n}} {(- 1)}^{m} \cdot^{n} C_{m} \frac{\sec^{2 m} θ}{2 m}]

l i m i t s 0 t o \frac{π}{4}

= {(- 1)}^{n} [\ln \sqrt{2} + \underset{m = 1}{\sum^{n}} {(- 1)}^{m} \cdot^{n} C_{m} \cdot \frac{2^{m} - 1}{2 m}]

more simplifications to be done .

Commented by prakash jain last updated on 02/Nov/15

Can you please recheck and confirm the

question . Because

^{n} C_{m} (2^{m} - 1) \neq 1

If the question is correct can you review

the integral in comment .

Commented by prakash jain last updated on 03/Nov/15

Q : {(- 1)}^{n} (\frac{1}{2} l n 2 + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{2 m})

What i got : {(- 1)}^{n} (\frac{1}{2} l n 2 + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}^{n} C_{m} (2^{m} - 1)}{2 m})

Red color is extra factor which needs to be

removed .

Commented by Yozzi last updated on 03/Nov/15

I d e r i v e d a r e d u c t i o n f o r m u l a w h i c h

e x c l u d e s t h e f a c t o r (\begin{matrix} n \\ m \end{matrix}) (2^{m} - 1)

w h e n u s e d t o p r o v e t h e a n s w e r .

B u t, I {d o n}^{'} t k n o w h o w t h e f a c t o r {(- 1)}^{n}

c o m e s i n .

I_{n} + I_{n - 1} = \frac{1}{2 n}

n ⩾ 2, I_{n} = \int_{0}^{π / 4} {t a n}^{2 n + 1} θ d θ .

Commented by prakash jain last updated on 03/Nov/15

I_{n} + I_{n - 1} = \frac{1}{2 n}

I_{n} = \frac{1}{2 n} - \frac{1}{2 (n - 1)} + \dots . + {(- 1)}^{n} \ln \sqrt{2}

I_{n} = {(- 1)}^{2 n} \frac{1}{2 n} + {(- 1)}^{2 n - 1} \frac{1}{2 (n - 1)} + \dots + {(- 1)}^{n + 1} \frac{1}{2} + {(- 1)}^{n} \ln \sqrt{2}

If you take {(- 1)}^{n} common you will get the

result in question using the formula .

Commented by Yozzi last updated on 03/Nov/15

I s e e . T h a n k s!

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