JEE Math practice: Indefinite Integral

Calculus Level 4

Let $f$ be a continuous function satisfying $f(x+y)=f(x)+f(y)$ , for each $x,y\in\mathbb{R}$ and $f(1)=2$ then $\int\frac{f(x)\tan^{-1}x}{(1+x^2)^2}dx$ is equal to

C-\frac{\tan^{-1}x}{2(1+x^2)}+\frac14\tan^{-1}x+\frac{x}{4(1+x^2)}

\text{cannot be determined explicitly}

C-\frac{\tan^{-1}x}{2(1+x^2)}+\frac14\tan^{-1}x+\frac{f(x)}{1+(f(x)^2)}

C-\frac1{1+x^2}\tan^{-1}x+\frac12\tan^{-1}x+\frac x{2(1+x^2)}

1 solution

Aditya Agarwal
Jan 5, 2016

$f(x+y)=f(x)+f(y)$ $f(1+x)=f(x)+2$ $\implies f(1+x)=f(x-1)+2+2$ Continuing like that, we get $f(1+x)=f(x-(x-1))+2(x-1+1)$ $\implies f(1+x)=f(1)+2(x)$ $\implies f(x)=2x$ So the integral becomes $\int\frac{2x\tan^{-1}x}{(1+x^2)^2}dx$ Putting $\tan^{-1}x=t$ , we get $\int 2t\tan t\cos^2x dx=\int t\sin 2tdt$ Which is easy to compute by Integration by parts or using the imaginary part of complex numbers.

Please explain how we get $f(x)=2x$ from the recurrence.

Otto Bretscher - 5 years, 5 months ago

Is it right? I was confused about that part. (I edited)

Aditya Agarwal - 5 years, 5 months ago

Your step "continuing like that" works for integers $x$ only. Hint: Prove $f(x)=2x$ for the rational numbers. Then, by continuity, $f(x)=2x$ for all real numbers.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Oh yes, I overlooked that. Thanks for pointing out. Can you give an insight on how can I do this?

Aditya Agarwal - 5 years, 5 months ago

@Aditya Agarwal – The way I think about it, it requires a few steps, but each step is straightforward. Show the following:

(I) $f(0)=0$

(II) $f(-x)=-f(x)$ for all real $x$

(III) $f(m)=2m$ for positive integers $m$

(IV) $f(m/n) = 2m/n$ for positive integers $n,m$ .

Now the equation $f(x)=2x$ holds for all real numbers $x$ , by continuity.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Okay, I got it thanks. But do I need to edit my answer? Because in JEE you really don't get that much time. In fact, there, you start by making an educated guess and differentiate it. (Reverse of what we are told to do)

Aditya Agarwal - 5 years, 5 months ago

@Aditya Agarwal – Your solution is not complete... in fact, it skips over the key issue.

Let me outline a solution for you:

(I) $f(0)=f(0+0)=f(0)+f(0)$ so $f(0)=0$

(II) $0=f(0)=f(x+(-x))=f(x)+f(-x)$ so $f(-x)=-f(x)$ for all real $x$

(III) $f(m)=f(1+1+...+1)=f(1)+f(1)+...+f(1)=mf(1)=2m$ for positive integers $m$

(IV) $2m=f(m)=f(\frac{m}{n}+\frac{m}{n}+...+\frac{m}{n})=f(\frac{m}{n})+f(\frac{m}{n})+...+f(\frac{m}{n})=nf(\frac{m}{n})$ so $f(\frac{m}{n})=\frac{2m}{n}$ for positive integers $m,n$ .

Now $f(x)=2x$ for all real $x$ by continuity; any real number is the limit of a sequence of rational numbers.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – But is it a coincidence that this function, is indeed correct, though I proved it only for integers? If so, how? Doesn't this mean we can extend the proof from here?

Aditya Agarwal - 5 years, 5 months ago

JEE Math practice: Indefinite Integral

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