5
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I want to prove that the following integral exists: $$ \int_0^\infty e^{-x}\left(1+\sin\left(x^2\right) \right) dx $$

First I tried to calculate it by splitting it up (multiplied it out and then separated it at the sum symbol) but then I got stuck.

Then I tried an other way. An improper Integral $[0,\infty)$ exists, if $0\leq \int{f(x)}\leq \int{g(x)}$, if $\int{g(x)}$ exists (obviously $f:=e^{-x}...$).

Since $|\sin(x^2)|\leq1$ therefore $$ 0 \le \int_0^\infty e^{-x}\left(1+\sin\left(x^2\right) \right) dx \le 2\int_{0}^{\infty}e^{-x}dx = 2, $$ so the improper integral exists and is less or equal then 2 (in fact it is approx. 1,27 I think).

Is this right or would you use another approach? Thank you in advance!

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  • $\begingroup$ looks good to me $\endgroup$ – gt6989b Feb 5 at 15:57
  • 1
    $\begingroup$ Exactly $\frac 75$ and your approach is good to me $\endgroup$ – Claude Leibovici Feb 5 at 15:57

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