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We are given $$\int^{\infty}_{3} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx = I_1 + I_2,$$ $ where$ $I_1 = \int^{4}_{3} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx $ and $I_2 = \int^{\infty}_{4} \frac{x}{(x-3)^{\frac{3}{2}}}\, \mathrm dx $

State whether $I_1$ and $I_2$ are convergent or divergent.

I tried comparing to $ \frac{1}{(x-3)^{\frac{3}{2}}} $ and other similar integrals but I couldn't get a solution. All help is appreciated, thanks in advance.

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Set $u=x-3$. The integral becomes $$\displaystyle\int^{\infty}_{3} \frac{x}{(x-3)^{3/2}}\, \mathrm dx = \int^{\infty}_{0} \frac{u+3}{u^{3/2}}\, \mathrm du = \underbrace{\int^{1}_{0} \frac{u+3}{u^{3/2}}\, \mathrm du}_{\mathstrut J_1}+\underbrace{\int^{\infty}_{10} \frac{u+3}{u^{3/2}}\, \mathrm du}_{\mathstrut J_2}$$

Use asymptotic analysis:

  • Now near $0$, $\dfrac{u+3}{u^{3/2}}\sim\dfrac{3}{u^{3/2}}$, so $J_1$, hence $I_1$, diverges.

  • Near $\infty$, $\dfrac{u+3}{u^{3/2}}\sim\dfrac{u}{u^{3/2}}=\dfrac{1}{\sqrt u}$, so $J_2$, hence $I_2$, diverges.

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Hint. Note that for $x\in (3,4]$ then $$ \frac{x}{(x-3)^{\frac{3}{2}}}\geq \frac{1}{(x-3)^{\frac{3}{2}}}\implies I_1\geq \int_3^4\frac{dx}{(x-3)^{\frac{3}{2}}}$$ Moreover for $x\geq 4$, $$ \frac{x}{(x-3)^{\frac{3}{2}}}\geq \frac{x}{(x)^{\frac{3}{2}}}=\frac{1}{x^{\frac{1}{2}}}\implies I_2\geq \int_4^{\infty}\frac{dx}{x^{\frac{1}{2}}}.$$ Can you take it from here?

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