Can anyone help to determine the domain of the function $f(x)$ for which the integral has a value of one?

$$\int_{-\infty}^{\infty}{1\over (\pi{x}+f(x))^2+1}dx=1\tag1$$

And how can one show $(1)$ has a value of 1 for any valid $f(x)$?

Setting $f(x)=\tan{x}$

$$\int_{-\infty}^{\infty}{1\over (\pi{x}+\tan{x})^2+1}dx=1\tag2$$

Which had been proven here

I have tried and set $f(x)$ to varies functions such as $\cos{x}$,$e^{x}$, $x\cos{x}$,... according to wolfram integrator it works fine.

I could try: enforcing $u=\pi{x}+f(x)$ then $du=\pi+f^{'}(x)dx$

$$\int_{-\infty}^{\infty}{1\over (1+x^2)(\pi+f^{'}(x))}du\tag3$$

which maybe of similiar to this post

  • $\begingroup$ I can hardly make sense out of your question. There isn't a single function such that the given integral equals one, and even assuming that, $\int_{-\infty}^{+\infty}g(x)\,dx = 1$ tells us very little about the domain of $f$. We may remove $x=1$, then $x=2$, then $x=3$, $\ldots$ from the domain of $f$ and the value of its integral is unchanged. $\endgroup$ – Jack D'Aurizio Jan 7 '17 at 20:28

Your statement is NOT true for general $f(x)$ in the slightest

Let $f(x) = x$. We then have $$\int_{-\infty}^{\infty}\frac{1}{(\pi x + x)^2+1}dx$$ $$=\int_{-\infty}^{\infty}\frac{1}{[x(\pi + 1)]^2+1}dx$$ $$=\int_{-\infty}^{\infty}\frac{1}{x^2(\pi+1)^2+1}dx$$ We now let $u=(1+\pi)x$ and thus $du = (1+\pi)dx$ $$=(1+\pi)\int_{-\infty}^{\infty}\frac{1}{u^2+1}du$$ The antiderivative is $\frac{\arctan(x+\pi x)}{1 + π}$, and when we take limits we get $$=\frac{\pi}{1+\pi}$$
Therefore, your statement is not true for all $f(x)$. It might be true for some specific $f(x)$ such as $f(x)=\tan(x)$, but I would not expect this to hold given an arbitrary function. Moreover, given that Mathematica fails to calculate the integral for $f(x)=\sin(x)$ or even $f(x)=e^x$ I am fairly sure the online integrator you were using is wrong. If by "domain where this holds" you meant the "class of functions for which this holds" then you might have something, but that is not what your post asks for at the moment

  • $\begingroup$ Thanks, ok. I try to repost it or ask a more sensible one. $\endgroup$ – user339807 Jan 7 '17 at 20:15

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