calculus - Evaluate $int_{0}^{infty}sqrt{frac{ a^2 -y^2 + sqrt{(a^2-y^2)^2+4y^2} }{(a^2-y^2)^2+4y^2}}dy=sqrt{2}pi$

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.

Answer

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.

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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 $y\to\frac\pi2$.