Functional Integration
f ( x ) + f ( 1 − x 1 ) = tan − 1 x , ∀ x ∈ R − { 0 }
If f : R → R satisfies the functional equation above then ∫ 0 1 f ( x ) d x = c a π b , where a , b , c are positive integers with a , c coprime.
Find the value of a + b + c .
The answer is 12.
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1 solution
I think there is a contradiction: Say x is in (0,1).You have shown that f(x)+f(1-x)=3π/4.
Let us take g(x)=f(x)+f(1-x). So, g(x)=3π/4. Now, given equation is f(x)+f(1-1/x)=arctan(x)
Replace x by 1/x. So, f(1/x)+f(1-x)=arctan(1/x).Now, add this equation with the equation given in question.
Then we get [f(x)+f(1-x)]+[f(1/x)+f(1-1/x)]=arctan(x) +arctan(1/x) That is, g(x)+g(1/x)=arctan(x)+arctan(1/x).
Now,g(x) is a constant function=3π/4.So,g(1/x)=3π/4.
Hence we get 3π/2= arctan(x)+arctan(1/x) which is not possible for any x.
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What makes you certain about g(1/x)=3 π /4 its domain is not (0,1).
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The argument I made works for all x>0, not necessarily for x belonging to (0,1). You can see that the argument I made his for all positive x.Assuming your result to be true, g(x) is a constant function which implies g(1/x) is 3π/4.
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@Indraneel Mukhopadhyaya – My result is true for x in the range specified, g(x) for x> 1 is -pi/4, you can check that by considering equations 4 and 5 in (1,infinity) as the last two terms reduce to -pi because of domain implications.
I will add details in the solution in a few hours.
@Indraneel Mukhopadhyaya
–
Here is the graph of g(x)
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@Kushal Patankar – Yes, I got it now. Thanks for the clarification. Everything is fine in your proof.
( link try to do this problem please
\begin{aligned} f(x)+f \left(1-\dfrac{1}{x}\right) & =\tan^{-1} (x) \tag{1} \\ \small\color{#3D99F6}{\text{Replace } x \text{ by } \dfrac{x-1}{x},\text{ we get}}\\ f \left(\dfrac{x-1}{x}\right)+f \left(\dfrac{1}{1-x}\right) & =\tan^{-1}\left(\dfrac{x-1}{x}\right)\tag{2} \\ \small\color{#3D99F6}{\text{Replace } x \text{ by } \dfrac{1}{1-x},\text{ we get}}\\ f \left(\dfrac{1}{1-x}\right)+f(x) & =\tan^{-1}\left(\dfrac{1}{1-x}\right)\tag{3} \\ \small\color{#3D99F6}{(1)-(2)+(3), \text{we get,}}\\ 2f(x) &= \tan^{-1} x + \tan^{-1} \left(\dfrac{1}{1-x}\right) -\tan^{-1}\left(\dfrac{x-1}{x}\right)\tag{4} \\ \small\color{#3D99F6}{\text{Replace } x \text{ by } (1-x),\text{ we get}} \\2f(1-x) &= \tan^{-1} (1-x) + \tan^{-1} \left(\dfrac{1}{x}\right) -\tan^{-1}\left(\dfrac{x}{x-1}\right)\tag{5} \\ \color{#3D99F6}{\text{Adding } eq^ns \space (4) \text{ and } (5)} \\ 2\left(f(x)+f(1-x)\right) & = \frac{3 \pi}{2} \tag{6} \\ f(x)+f(1-x)& = \frac{3 \pi}{4} \tag{7} \\ \color{#3D99F6}{\text{Now,}} & \\ \displaystyle\int_{0}^{1} f(x) \space \text{d}x &= \displaystyle\int_{0}^{1} f(1-x) = \displaystyle\int_{0}^{1} \frac{1}{2}\left(f(x)+f(1-x)\right) = \frac{3 \pi}{8}\\ \frac{3 \pi}{8} &=\frac{a \pi ^{b}}{c}, \text{ thus,} \\ a+b+c &= 3+8+1 =\boxed{12} \end{aligned}
Edit :
\begin{aligned} 2f(x) + 2f(1-x) &= \tan^{-1} x + \tan^{-1} \left(\dfrac{1}{1-x}\right) -\tan^{-1}\left(\dfrac{x-1}{x}\right) +\tan^{-1} (1-x) + \tan^{-1}\left(\dfrac{1}{x}\right) -\tan^{-1}\left(\dfrac{x}{x-1}\right)\quad \quad \quad \quad\tag{5.1}\\ 2(f(x) + f(1-x))&= \left( \tan^{-1} x + \tan^{-1}\left(\dfrac{1}{x}\right) \right)+\left(\tan^{-1} (1-x)+\tan^{-1} \left(\dfrac{1}{1-x}\right) \right)-\left(\tan^{-1}\left(\dfrac{x-1}{x}\right)+\tan^{-1}\left(\dfrac{x}{x-1}\right)\right)\tag{5.2} \end{aligned}
tan − 1 x + tan − 1 ( x 1 ) tan − 1 ( 1 − x ) + tan − 1 ( 1 − x 1 ) tan − 1 ( x x − 1 ) + tan − 1 ( x − 1 x ) = = = 2 π , 2 π , − 2 π , ∀ x ∈ ( 0 , 1 ) ∀ x ∈ ( 0 , 1 ) ∀ x ∈ ( 0 , 1 )