# How to calculate $\lim\limits_{x\to1^+}\frac{1}{(x-1)^2} \int\limits_{1}^{x} \sqrt{1+\cos(\pi t)}\,\mathrm dt$

Can anyone help me by calculating this limit?

I know that I need L'Hôpital but how?

$$\lim_{x \to 1^+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+\cos(\pi t)} \,\mathrm dt$$

Thank you very much!!

• Please use different symbols for the upper bound of the integral and for the integrand. – Did May 21 '13 at 15:31

Since my hints were not helpful, here a is my solution \begin{align} \lim\limits_{x\to 1^+}\frac{1}{(x-1)^2}\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt &=\lim\limits_{x\to 1^+}\frac{\left(\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt\right)'}{((x-1)^2)'}&\text{L'Hopitale rule}\\ &=\lim\limits_{x\to 1^+}\frac{\sqrt{1+\cos(\pi x)}}{2(x-1)}& y\to x-1\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{1+\cos(\pi (y+1))}}{2y}&\cos(\pi+\alpha)=-\cos\alpha\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{1-\cos(\pi y)}}{2y}&1-\cos\alpha=2\sin^2\frac{\alpha}{2}\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}|\sin\frac{\pi y}{2}|}{2y}&y>0\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\sin\frac{\pi y}{2}}{2y}&\sin\alpha\sim\alpha\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\frac{\pi y}{2}}{2y}\\ &=\frac{\pi\sqrt{2}}{4} \end{align}
• $$\lim_{x,1+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+cos(\pi t)} dt$$ i used L'Hospital so i got this: $$\frac{\sqrt{1+cos(\pi *x)}} {2*(x-1)}$$ so i used l'Hospital again: and again...and again so i got this: $$\frac{\pi * (2*cos(\pi*x))+(3*cos(2*\pi *x)+2))*\sqrt{1+cos(\pi*x)}} {2*sin(\pi*x)*(3*cos(\pi*x)+2)}$$ now i dont know how to continue. – Xorkox May 21 '13 at 15:55
• You need to aplly that rule only once, then substitute $y=x-1$, then $y\to 0$ – Norbert May 21 '13 at 16:04
Integrate the equivalent, valid when $t\gt1$, $t\to1$, $$\sqrt{1+\cos(\pi t)}=\sqrt2\,\sin\left(\frac\pi2(t-1)\right)\sim\frac\pi{\sqrt2}\,(t-1).$$ This yields, for $x\to1$, $x\gt1$, $$\int_1^x\sqrt{1+\cos(\pi t)}\mathrm dt\sim\frac\pi{\sqrt2}\int_1^x(t-1)\mathrm dt=\frac\pi{2\sqrt2}(x-1)^2.$$