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I need to check if the following integral converges / diverges conditionally. $$\int_0^\infty x\sin(x^3) dx$$

I have tried integrating by parts and it didn't work. Will appreciate any help :).

Sumanta
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Shahar Ziv
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  • Tried that as well. Didn't work for me unfortunately. – Shahar Ziv Jul 09 '20 at 08:55
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    @ShaharZiv It would be better if you'd write all that you tried in your question – Arnav Mahajan Jul 09 '20 at 08:56
  • Just a question I am not familiar with convergence and divergence however is it necessary to compute the integral itself to check for those? The integral in itself seems to have no simple solution and the answer uses functions such as Li – FoundABetterName Jul 09 '20 at 08:58
  • there are several convergene test for improper integral suvh as Abel's or Dirichlet's test which do not involve directly calculating the integrals. – Shahar Ziv Jul 09 '20 at 09:09

3 Answers3

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The convergence can be based on the alternating series test.

Let $0 < \alpha < \beta$. Then

$$ \int_{\alpha}^{\beta} x \sin(x^3) \, dx = \int_{\alpha^3}^{\beta^3} t^{1/3} \sin(t) \cdot \frac13 t^{-2/3} \, dt = \frac13 \int_{\alpha^3}^{\beta^3} t^{-1/3} \sin(t)\, dt $$ by change of variables, so your integral converges if and only if $$ \int_{2\pi}^{\infty} t^{-1/3} \sin(t)\, dt $$ converges; it does since the subsequent integrals

$$ \int_{2\pi}^{3\pi} t^{-1/3} \sin(t)\, dt, \int_{3\pi}^{4\pi} t^{-1/3} \sin(t)\, dt, \cdots $$ form an alternating sequence which decreases monotonically in absolute value and converges to zero.

Steven
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Ok so applying Yves' suggestion I get: $\int_0^\infty x\sin(x^3) dx = \int_0^\infty\sin(t)/t^{-3}dt$ which converges. Thank you!

Just_A_User
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Shahar Ziv
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  • I think you might be missing a factor of $3$ somewhere. Also, to make this more useful for future readers, you could add some more information as to why the second integral converges! :) – Aryaman Maithani Jul 09 '20 at 09:51
  • Sorry, your change of variable is wrong. –  Jul 09 '20 at 12:20
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Hint: You can start from

$$\sin x=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}$$ hence your integral can be written like:

$$\int_0^\infty x\sin(x^3) dx=\int_0^\infty\dfrac{ix\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)}{2i^2}\,\mathrm{d}x=\int_0^\infty-\dfrac{ix\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)}{2}\,\mathrm{d}x$$

$$=\frac i2\int_0^\infty x\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)\,\mathrm{d}x$$ Last integral is simple to solve using the substitution $t=-x^2 \iff \mathrm{d}x=-\frac{1}{2x}\,\mathrm{d}t$ for $\int_0^\infty x\mathrm{e}^{ix^3}\,\mathrm{d}x$.

It is necessary to remember incomplete gamma function https://en.wikipedia.org/wiki/Incomplete_gamma_function

Sebastiano
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