Find $k\in\Bbb R$ if $\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=\int_0^\infty xe^{-x}\,\mathrm dx$

Question

Find $k\in\Bbb R$ if $$\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=\int_0^\infty xe^{-x}\,\mathrm dx.$$

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$$\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=\lim_{x\to0}\frac{k\sin(kx)}{2x}=\lim_{x\to0}\frac{k^2}{2}\frac{\sin(kx)}{kx}=\frac{k^2}2.$$ To find $\int_0^\infty xe^{-x}\,\mathrm dx$ we can add a limit: $$\lim_{b\to\infty}\int_0^bxe^{-x}\,\mathrm dx.$$ Now we can integrate by parts. Thus $u=x$, so $u'=1$, and $\mathrm dv=e^{-x}$, so $v=-e^{-x}$, thus $$\int xe^{-x}\,\mathrm dx=uv-\int u'v\,\mathrm dx=-xe^{-x}+\int e^{-x}\,\mathrm dx=-xe^{-x}-e^{-x}=-e^{-x}(x+1),$$ hence $$\lim_{b\to\infty}\left[-e^{-x}(x+1)\right]_0^b=-\lim_{b\to\infty}(e^{-b}(b+1)-1)=-(0-1)=1.$$ Hence, $$\frac{k^2}2=1\implies\boxed{k=\sqrt2\vee k=-\sqrt2}.$$ Is it correct?

Thanks!!

Answer 1

Yes, this is correct. You could vary certain details. For example, $1-\cos kx=2\sin^2\frac{kx}{2}$ would let you evaluate the left-hand side without any calculus, if you know the famous geometric result $\frac{\sin x}{x}\sim 1$.