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Evaluate $$I=\int_0^{\infty}\frac{1}{(1+x^9)(1+x^2)}dx$$

I substituted $x=\tan y$ which gives $dx=\sec^2ydy$.

Now I am having problem in determining the limits of new integral formed after substitution because $\tan y=0$ and $\lim_{x\to y}\tan y=\infty$ have infinetly many solutions for $y$ which on substituting and then evaluating the integral using following propery$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx $$ gives different solutions which is not possible.

Where is the trouble? I think there is some problem in substitution.

Lalit Tolani
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1 Answers1

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The substitution bijects for $y\in[0,\,\pi/2)$. The definition$$\begin{align}\int_0^\infty\frac{f(x)dx}{1+x^2}&:=\lim_{X\to\infty}\int_0^X\frac{f(x)dx}{1+x^2}\\&=\lim_{X\to\infty}\int_0^{\arctan X}f(\tan y)dy\\&=\int_0^{\lim_{X\to\infty}\arctan X}f(\tan y)dy\\&=\int_0^{\pi/2}f(\tan y)dy\end{align}$$solves your problem. I'm confident you can use this to prove more generally$$\int_0^\infty\frac{dx}{(1+x^k)(1+x^2)}=\frac{\pi}{4}.$$

J.G.
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  • But $x=\tan y$ does not implies $y=\arctan x$ – Lalit Tolani Dec 26 '21 at 19:05
  • @Jean-ClaudeArbaut So basically what we do is substitute x such that $y=\arctan x$ – Lalit Tolani Dec 26 '21 at 19:29
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    @LalitTolani Sorry, got it wrong. In the formula here, you have $\varphi(u)=\tan(u)$. What count here is $\varphi$ must be continuous and differentiable (and you have to find the original bounds with $\varphi(a)$ and $\varphi(b)$). Hence there are other possibilities for the bounds, but the simplest is $(0,\pi/2)$. If you add $k\pi$ to both, it's still correct. – Jean-Claude Arbaut Dec 26 '21 at 19:37
  • @Jean-ClaudeArbaut , Clear now, the thing I was missing that $\phi$ must be differentiable – Lalit Tolani Dec 26 '21 at 19:48