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Prove that $$\lim_{\lambda\rightarrow\infty}\int_1^2\frac{\cos\lambda t}{t\sqrt{t-1}}\text{d}t=0.$$

I have tried differentiating the integrand w.r.t $\lambda$ but it doesn't look promising.

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    $\begingroup$ Have you already encountered the Riemann-Lebesgue lemma? $\endgroup$ – Daniel Fischer Jun 24 '14 at 19:17
  • $\begingroup$ Hint. $\lim \int_a^b f(t) \cos \lambda t dt = 0$ for all sufficiently nice functions $f$. $\endgroup$ – Karolis Juodelė Jun 24 '14 at 19:19
  • $\begingroup$ @Karolis It looks like a fourier coefficient and then? $\endgroup$ – lovelesswang Jun 24 '14 at 19:31
  • $\begingroup$ @lovelesswang, it might be simpler to think about Ryman sums of $f$ and take $\Delta x = \frac {2 \pi}{\lambda}$. $\endgroup$ – Karolis Juodelė Jun 24 '14 at 20:03
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$$\begin{eqnarray*}\int_{1}^{2}\frac{\cos(\lambda t)}{t\sqrt{t-1}}\,dt &=&\cos(\lambda)\int_{0}^{1}\frac{\cos(\lambda t)}{(t+1)\sqrt{t}}-\sin(\lambda)\int_{0}^{1}\frac{\sin(\lambda t)}{(t+1)\sqrt{t}}\\&=&2\cos(\lambda)\int_{0}^{1}\frac{\cos(\lambda u^2)}{1+u^2}\,du-2\sin(\lambda)\int_{0}^{1}\frac{\sin(\lambda u^2)}{1+u^2}\,du\tag{1}\end{eqnarray*}$$ where both $\sin(\lambda),\cos(\lambda)$ are bounded by $1$ in absolute value. $$\int_{0}^{1}\frac{\sin(\lambda u^2)}{1+u^2}\,du = \frac{1}{\sqrt{\lambda}}\int_{0}^{\sqrt{\lambda}}\frac{\sin v}{1+\frac{v^2}{\lambda}}=\frac{1-\cos(\sqrt{\lambda})}{2\sqrt{\lambda}}+\frac{1}{\sqrt{\lambda}}\int_{0}^{\sqrt{\lambda}}\frac{2v(1-\cos v)}{\lambda\left(1+\frac{v^2}{\lambda}\right)^2}\,dv\tag{2}$$ by integration by parts, where the last integral is non-negative and bounded by: $$ \frac{1}{\sqrt{\lambda}}\int_{0}^{\sqrt{\lambda}}\frac{4v\,dv}{\lambda(1+\frac{v^2}{\lambda})^2}=\frac{1}{\sqrt{\lambda}}\tag{3}$$ and the same argument applies to the integral depending on $\cos(\lambda u^2)$.
By putting all together, we get:

$$ \int_{1}^{2}\frac{\cos(\lambda t)}{(t+1)\sqrt{t}}\,dt = O\left(\frac{1}{\sqrt{\lambda}}\right)\tag{4} $$

as $\lambda\to +\infty$.

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