Evaluate $int_{0}^infty frac{sin(x)}{sqrt{x^2+1}} dx$

Question

The first substitution that came to my mind was
$$
begin{cases}
x = tan(theta) \
dx = sec^2theta dtheta \
textrm{Bounds:} , , 0 to , pi/2
end{cases}
$$
We now have:
$$
int_0^{frac{pi}{2}}sin(tan(theta))cos^4theta , d theta
$$
But that really doesn't seem to help. Any advice is greatly appreciated.

J.G. · Answer

Changing the integration variable to $t$ for ease of comparison with Eq. 11.5.7 in @metamorphy's DLMF link, $nu=0,,x=1$ gives$$int_0^infty(1+t^2)^{-1/2}sin tmathrm dt=frac{pi}{2}(I_0(1)-L_0(1)),$$with $I_0$ a modified Bessel function of the first kind and $L_0$ a modified Struve function.

Answered by J.G. on February 20, 2021

Evaluate $int_{0}^infty frac{sin(x)}{sqrt{x^2+1}} dx$

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