# Proof that improper integral $\int_{0}^{\infty} e^{-x}(1+\sin(x^2)) dx$ exists

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!

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