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Let $F,G$ be defined by: $$F(x)=\int_{0}^{x} e^{-t^2}\mathrm dt$$$$G(x)=\int_{0}^{1}\dfrac{e^{-x^2(t^2+1)}}{t^2+1}\mathrm dt$$ Prove that $F(x)^2+G(x)=\frac{\pi}{4}$

I can prove that $F(x),G(x)$ are derivable. And: $$F'(x)=e^{-x^2}$$$$G'(x)=\int_{0}^{1}-2x(t^2+1)e^{-x^2(t^2+1)}\mathrm dt=\int_{0}^{-2x^2}e^u\mathrm du=e^{-2x^2}-1$$, with $u=-x^2(t^2+1)$

And then I try to prove that $2F'(x)F(x)+G'(x)=0$

But I can't finish my proof.

liverpool29
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2 Answers2

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hint

$$2F'(x)F (x)+G'(x)=0 \implies $$ $$F^2 (x)+G (x)=constant=$$ $$=F^2 (0)+G (0) $$

$$=G (0)=\int_0^1\frac {dt}{1+t^2} $$ $$=\arctan (1)=\frac {\pi}{4} $$

hamam_Abdallah
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  • Thank you but can you give some hints to prove that $F(x)^2+G(x)=constant$? – liverpool29 Sep 22 '17 at 14:24
  • Why this sanction. what is wrong. – hamam_Abdallah Sep 22 '17 at 14:25
  • I didn't put -1 for your answer, I just don't know how to prove $F(x)^2+G(x)=cst$ – liverpool29 Sep 22 '17 at 14:27
  • I upvoted, it looks fine to me, I was about to write a similar answer. It is easy to prove that F(x)^2+G(x)=c from the first line, because the first line is the derivative of that, which is zero. To prove the first line, just sub in the formulas from the question and compute. – Teddy Baker Sep 22 '17 at 14:27
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Two problems here: your differentiation of $G$ is wrong, and the change of variables doesn't work like that (you have to differentiate with respect to $t$ when changing variables, not $x$, which is constant inside the integral).

You should have $$ G'(x) = \int_0^1 -2xe^{-x^2(1+t^2)} dt, $$ and setting $t=xu$, $dt=x\,du$, so $$ F(x) = \int_0^1 xe^{-x^2u^2} du. $$ Then, since $F'(x)=e^{-x^2}$, we have $$ 2F'(x)F(x) = 2e^{-x^2}\int_0^1 xe^{-x^2u^2} du = \int_0^1 2xe^{-x^2(1+t^2)} dt = -G'(x) $$ changing the label on the integration variable in $F$.

Chappers
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  • Thank you but I don't understand why G'(x) is diffenrentiate wrt t. I think that this link shows that it should be x https://www.encyclopediaofmath.org/index.php/Parameter-dependent_integral – liverpool29 Sep 22 '17 at 14:36
  • Oh, I see, there's somthing wrong in my calculate. Thanks again – liverpool29 Sep 22 '17 at 14:39