2
$\begingroup$

I want to compute $$\lim_{t \to \infty} \int_1^2 \frac{\sin (tx)}{x^{2}(x-1)^{1/2}}\,dx. $$

The integrand has discontinuity at $x=1$, so the integral is equal to the following limit: $$\lim_{t \to \infty}\lim_{s \to 1^+} \int_s^2 \frac{\sin (tx)}{x^{2}(x-1)^{1/2}}dx, $$ and I use substitution $tx= a$; then $tdx=da$.

$$\lim_{t \to \infty}t^{3/2}\lim_{s \to 1^+}\int_{st}^{2t} \frac{\sin (a)}{a^{2}(a-t)^{1/2}}da $$

how to proceed this integral?

  • 0
    Do you mean $\dfrac{\sin(tx)}{x^2(x-1)^{1/2}}$ or $\left(\dfrac{\sin(tx)}{x^2}\right)(x-2)^{1/2}$ ?2012-03-09
  • 0
    i mean the first one2012-03-09

1 Answers 1