Let $\phi(z)$ be analytic function defined on the half-plane $$H(\delta)=\{z\in \mathbb{C}: \operatorname{Re}z\ge -\delta\}$$ for $0<\delta<1$. Suppose that, for some $A<\pi$, $\phi$ satisfies the growth condition $$|\phi(x+iy)|\le Ce^{Px+ A|y|}$$ for all $z\in H(\delta)$. Then, for any $0<\operatorname{Re} s< \delta$, Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$ as long as the hypotheses of the theorem are satisfied.
Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$.
Let $\phi(z)$ be analytic function defined on the half-plane $$H(\delta)=\{z\in \mathbb{C}: \operatorname{Re}z\ge -\delta\}$$ for $0<\delta<1$. Suppose that, for some $A<\pi$, $\phi$ satisfies the growth condition $$|\phi(x+iy)|\le Ce^{Px+ A|y|}$$ for all $z\in H(\delta)$. Then, for any $0<\operatorname{Re} s< \delta$, Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$ as long as the hypotheses of the theorem are satisfied.
Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f(x^2)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$$$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f(x^2)\,dx=\frac{\pi}{2}\phi(-s). $$$$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$.
Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f(x^2)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f(x^2)\,dx=\frac{\pi}{2}\phi(-s). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$.
Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$.
Ramanujan's master theorem tells you that if $$f(x)=\sum_{k=0}^\infty \phi(k) (-x)^k,$$ then $$\int_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$ Therefore, $$\int_0^\infty x^{2s-1} f(x^2)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$ so that, letting $s=1/2$, $$\int_0^\infty f(x^2)\,dx=\frac{\pi}{2}\phi(-s). $$ The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$.