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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?

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Do you mean $\dfrac{\sin(tx)}{x^2(x-1)^{1/2}}$ or $\left(\dfrac{\sin(tx)}{x^2}\right)(x-2)^{1/2}$ ? –  Arturo Magidin Mar 9 '12 at 20:16
    
i mean the first one –  rose Mar 9 '12 at 20:18
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1 Answer

Do you know the Riemann-Lebesgue lemma?

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no but i am looking it right now –  rose Mar 9 '12 at 20:20
    
Follows quite trivially from this lemma. Good job, +1! –  Patrick Da Silva Mar 9 '12 at 20:27
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