Show that $\int_{0}^{1}\sin(\sinh(z)){\mathrm dz\over \sin(z^2)}=\gamma-\Gamma(\gamma)$

$$\int_{0}^{1}\sin(\sinh(z)){\mathrm dz\over \sin(z^2)}\tag1$$

we have shown that

$v=\sin(\sinh(z))$

$$\int_{0}^{\infty}\cos(\sinh(v)){\mathrm dv\over 1+v^2}\tag2$$

using the Feynman trick

we got to

$$\int_{0}^{\infty}x^{1\over \sin(z^2)}\tanh(v)\tag3$$

I am sure where to go from here, but I suspect that the closed form could be $\gamma-\Gamma(\gamma)$

• The integrand is $1/z + O(1)$ as $z \to 0$, so $\int_0^1 \frac{\sin(\sinh(z))}{\sin(z^2)}\; dz$ diverges. – Robert Israel Nov 13 '17 at 6:29
• The differential in integral $(3)$ seems to have been accidentally omitted – Chase Ryan Taylor Nov 13 '17 at 6:29
• There is smothing wrong somewhere (beside what Robert Israel commented). For $0 \leq z \leq 1$, function $\frac{\sin(\sinh(z))}{\sin(z^2)}$ is always positive while $\gamma-\Gamma(\gamma)\approx -0.966666$. Any typo ? – Claude Leibovici Nov 13 '17 at 7:39

The integral $\int_{0}^{1}\sin(\sinh z)\frac{dz}{\sin(z^2)}$ does not convergence because the integrand function behaves like $\frac{1}{z}$ in a right neighbourhood of the origin. $(2)$ is a different integral.