# How to evaluate an improper integral.

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.

## 1 Answer

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: $$I = \left(\sin(1)+\cos(1)\right)\sqrt{\frac{\pi}{2}}.\tag{3}$$

• I haven't known the Fresnel integral. Thank you! – user Sep 23 '15 at 11:35
• @P.Mike: Fresnel integrals are related to the Gaussian integral by way of Euler's formula. – Lucian Sep 24 '15 at 3:35
• @LucianThak you for giving nice information! I'll check later. – user Sep 24 '15 at 3:58