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.

  • 3
    $\begingroup$ 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

$$\begin{eqnarray*}\int_{1}^{2}\frac{\cos(\lambda t)}{t\sqrt{t-1}}\,dt &=&\cos(\lambda)\int_{0}^{1}\frac{\cos(\lambda t)}{(t+1)\sqrt{t}}-\sin(\lambda)\int_{0}^{1}\frac{\sin(\lambda t)}{(t+1)\sqrt{t}}\\&=&2\cos(\lambda)\int_{0}^{1}\frac{\cos(\lambda u^2)}{1+u^2}\,du-2\sin(\lambda)\int_{0}^{1}\frac{\sin(\lambda u^2)}{1+u^2}\,du\tag{1}\end{eqnarray*}$$ 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 $\lambda\to +\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.