How to evaluate an improper integral.

Question

I don't conceive how to evaluate the improper integral: $$ \int_{1}^{\infty}\frac{\sin x}{\sqrt{x-1}}dx. $$

When this converges, I'm glad if you give the value of this integral.

Thank you.

Answer 1

We have: $$ I = \int_{0}^{+\infty}\frac{\sin(x+1)}{\sqrt{x}}\,dx = \sin(1)\int_{0}^{+\infty}\frac{\cos(x)}{\sqrt{x}}\,dx+\cos(1)\int_{0}^{+\infty}\frac{\sin(x)}{\sqrt{x}}\,dx \tag{1}$$ and through a change of variable we get the Fresnel integrals: $$ \int_{0}^{+\infty}\frac{\cos(x)}{\sqrt{x}}=2\int_{0}^{+\infty}\cos(x^2)\,dx=\sqrt{\frac{\pi}{2}},$$ $$ \int_{0}^{+\infty}\frac{\sin(x)}{\sqrt{x}}=2\int_{0}^{+\infty}\sin(x^2)\,dx=\sqrt{\frac{\pi}{2}},\tag{2}$$ hence: $$ I = \left(\sin(1)+\cos(1)\right)\sqrt{\frac{\pi}{2}}.\tag{3} $$