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Prove or disprove that for any $a>1$,
$$\int_{0}^{\infty}\sqrt{\frac{ a^2 -y^2 + \sqrt{(a^2-y^2)^2+4y^2} }{(a^2-y^2)^2+4y^2}}dy=\sqrt{2}\pi~.$$

I came across with this integral evaluating inverse Laplace transformation of $\frac{1}{\sqrt{1+s^2}}$ without using any complex integral method. If possible I want to know if there's a way to show that $$\lim_{T\to\infty}\int_{a-iT}^{a+iT}{ds\over\sqrt{1+s^2}}=2\pi i, a>0$$ without using the residue method.

Actually I met a problem here by the reason explained in this question. Any help appreciated.

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  • $\begingroup$ Something to adjust in view of $$ restart; int(eval(sqrt((sqrt((a^2-y^2)^2+4*y^2)+a^2-y^2)/((a^2-y^2)^2+4*y^2)), a = 2), y = 0 .. infinity, numeric)$$ $$2.221441469 $$ and $$evalf(Pi/sqrt(2)) $$ $$ 2.221441469 .$$ $\endgroup$ – user64494 Jun 11 '14 at 16:38
  • $\begingroup$ @user64494 Thank you for your numerical evaluation. Now then, the integral must be $\sqrt{\pi\over 2}$, but then I have a contradiction that is mentioned in the linked question... $\endgroup$ – generic properties Jun 11 '14 at 16:41
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How do you get that a real integral over a real interval will result in an imaginary value?

Use $s=\sinh(t)$ to get $$ \int\frac1{\sqrt{1+s^2}}=\int \frac{\cosh(t)}{\sqrt{1+\sinh^2(t)}}dt=t+C $$ to evaluate this integral.


As to the changed boundaries, you still have to evaluate $$\operatorname{Arsinh}(a+iT)-\operatorname{Arsinh}(a-iT)=2i\,Im(\operatorname{Arsinh}(a+iT)).$$

Since $$ \sinh(x+iy)=\cos(y)\sinh(x)+i\sin(y)\cosh(x) $$ one has to solve $\cos(y)\sinh(x)=a$, $\sin(y)\cosh(x)=T$ for $y$. Combined, the equation $$ T=\sin(y)\sqrt{1+\frac{a^2}{\cos^2(y)}} $$ has to be solved in such a way that the solution $y(T)$ is continuous in $T$ from $0$ to $\infty$. Since $\sin(y)$ is bounded, large values for $T$ have to be reached by $\cos(y)\approx 0$, which means that $y\in[0,\frac\pi2)$ and the limit for $T\to\infty$ is reached by $y\to\frac\pi2$.

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  • $\begingroup$ Oops, I'm sorry. It was a typo. The integral is over an imaginary axis translated by some positive constant. $\endgroup$ – generic properties Jun 11 '14 at 17:10
  • $\begingroup$ Still the same substitution applies, you just have to make sure that the branches of the complex variant of $\operatorname{Arsinh}(s)$ connect continuously on the line of integration. $\endgroup$ – LutzL Jun 11 '14 at 17:15

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