[Integration #1] tan ( x f ( x ) ) = x 1 \tan(xf(x))=x-1

Calculus Level 5

A function f : R { 0 } R f : \mathbb{R}-\{0\} \to \mathbb{R} satisfies

{ tan ( x f ( x ) ) = x 1 f ( x ) < π 2 x \begin{cases} \tan (xf(x))=x-1 \\ \\ |f(x)|<\left|\dfrac{\pi}{2x}\right|\end{cases}

for all nonzero reals x x . The average of f ( x ) f(x) when x x is uniformly chosen from the interval [ 1 , 2 ] [1, 2] is p π q ln r + s p\pi^q\ln r + s for nonnegative rational numbers p p , q q , r r , and s s . Find p + q + r + s p+q+r+s to 4 decimal places.


The answer is 3.125.

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1 solution

Boi (보이)
Jun 9, 2018

Multiply x |x| to the inequality to obtain x f ( x ) < π 2 . |xf(x)| < \dfrac{\pi}{2}.

This means we can rewrite f ( x ) f(x) as f ( x ) = arctan ( x 1 ) x . f(x)=\dfrac{\arctan(x-1)}{x}.

The average value of f ( x ) f(x) over [ 1 , 2 ] [1,~2] is 1 2 arctan ( x 1 ) x d x . \displaystyle\int_1^2 \dfrac{\arctan(x-1)}{x}dx.


Use integration by parts, differentiating arctan ( x 1 ) \arctan(x-1) and integrating 1 x . \dfrac{1}{x}.

1 2 arctan ( x 1 ) x d x = [ ln x arctan ( x 1 ) ] 1 2 1 2 ln x ( x 1 ) 2 + 1 d x ( x 1 = tan t ) = π ln 2 4 0 π / 4 ln ( 1 + tan t ) d t \int_1^2 \dfrac{\arctan(x-1)}{x}dx \\ =\left[\ln x\cdot\arctan(x-1)\right]_1^2 - \int_1^2 \dfrac{\ln x}{(x-1)^2+1}dx ~{\small\color{#3D99F6} (\Longrightarrow~x-1=\tan t)} \\ =\frac{\pi\ln 2}{4} - \int_0^{\pi/4} \ln(1+\tan t)dt

Now, substitute t = π 4 u . t=\dfrac{\pi}{4}-u.

0 π / 4 ln ( 1 + tan t ) d t = 0 π / 4 ln ( 1 + tan ( π 4 u ) ) d u = 0 π / 4 ln ( 1 + 1 tan u 1 + tan u ) d u = π ln 2 4 0 π / 4 ln ( 1 + tan u ) d u \int_0^{\pi/4} \ln(1+\tan t)dt \\ = \int_0^{\pi/4} \ln\left(1+\tan\left(\dfrac{\pi}{4}-u\right)\right)du \\ =\int_0^{\pi/4} \ln\left(1+\dfrac{1-\tan u}{1+\tan u}\right)du \\ =\frac{\pi\ln 2}{4} - \int_0^{\pi/4} \ln(1+\tan u)du

Hence, 0 π / 4 ln ( 1 + tan t ) d t = π ln 2 8 . \displaystyle \int_0^{\pi/4} \ln(1+\tan t)dt = \dfrac{\pi\ln2}{8}.

From this we conclude that

1 2 arctan ( x 1 ) x d x = π ln 2 4 π ln 2 8 = π ln 2 8 . \int_1^2 \dfrac{\arctan(x-1)}{x}dx = \frac{\pi\ln2}{4}-\frac{\pi\ln2}{8}=\frac{\pi\ln2}{8}.

Then, p = 1 8 , q = 1 , r = 2 , s = 0. p=\dfrac{1}{8},~q=1,~r=2,~s=0.

p + q + r + s = 3.1250 . \therefore~p+q+r+s=\boxed{3.1250}.

0 0 isn't a positive rational number.

D G - 3 years ago

0 is not positive

Ραμών Αδάλια - 2 years, 12 months ago

Super sol and sums (+1)...But sir give the answer as 12.0000 since u said upto 4 places of decimal and it is accepting only integers..XD!!!!

rajdeep brahma - 2 years, 12 months ago

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