# $\lim_{n \to \infty} \int_{0}^{1} e^{x^2} \sin(nx) dx$ [duplicate]

$$\lim_{n \to \infty} \int_{0}^{1} e^{x^2} \sin(nx) dx$$ I am trying to find above limit without using the concept of Lebesgue integration. I know that $$e^{x^2} sin(nx)$$ is not a uniformly convergent sequence of functions so i can not take limit inside integration. So give me any idea that is it really possible to evaluate the limit without using the concept of lebesgue integration. Thank you

• Hint: let $z=nx$. May 6, 2021 at 20:26
• See this post. May 6, 2021 at 20:36
• Try Riemann-Lebesgue lemma. May 6, 2021 at 20:37

You integrate by parts: $$\int^1_0 e^{x^2} \sin(nx)~\mathrm{d}x = - \frac{1}{n}(e \cos(n) -1) + \frac{1}{n} \int^1_0 2x e^{x^2}\cos(nx)~\mathrm{d}x$$ Therefore, by using the triangle inequality, $$\lvert \cos(nx) \rvert \leq 1$$ and $$\left \lvert 2xe^{x^2} \right \rvert\leq 2e$$ on the domain: $$\left \lvert \int^1_0 e^{x^2} \sin(nx)~\mathrm{d}x \right \rvert \leq \frac{1}{n}\lvert e\cos(n) - 1\rvert + \frac{1}{n}\int^1_0 \left \lvert 2x e^{x^2} \right \rvert\lvert \cos(nx) \rvert~\mathrm{d}x \leq \frac{e-1}{n} + \frac{2e}{n}$$ The right hand side tends to zero when $$n \rightarrow \infty$$ and thus so does your integral.
In general: If $$f \in C^1([0, 1])$$, then $$\int^1_0 f(x) \sin(nx)~\mathrm{d}x \overset{n \rightarrow \infty}{\longrightarrow}0.$$ To prove this, you can use the same reasoning as presented above.
We can even allow $$f \in L^p([0, 1])$$ where $$1 \leq p \leq \infty$$. Because of density, for arbitrary $$\varepsilon > 0$$, we can choose $$g \in C^1_0([0, 1])$$ such that $$\lVert f - g \rVert_{L^1([0, 1])} < \frac{\varepsilon}{2}$$. Then: $$\left \lvert \int^1_0 f(x) \sin(nx)~\mathrm{d}x \right \rvert \leq \lVert f - g \rVert_{L^1([0, 1])} + \left \lvert \int_0^1 g(x)\sin(nx)~\mathrm{d}x \right \rvert < \frac{\varepsilon}{2} + \left \lvert \int_0^1 g(x)\sin(nx)~\mathrm{d}x \right \rvert$$ With the usual integration by parts strategy, we can make $$\displaystyle \left \lvert \int_0^1 g(x)\sin(nx)~\mathrm{d}x \right \rvert$$ smaller than $$\frac{\varepsilon}{2}$$.
• The general statement is also valid for any integrable function in $\mathbb{R}$. It is Féjer's formula. May 6, 2021 at 20:56
• @Meowdog Just to clarify what Oliver means is that $f$ need only be $C^0$, not $C^1$. May 6, 2021 at 23:13
• I reckon it even holds for $L^1$... I might have to make an edit... May 7, 2021 at 7:12
These are the Fourier coefficents of the function $$x \mapsto e^{x^2}I_{[0,1]}(x)$$. Invoke Riemann-Lebesgue: these converge to zero. Like the Ginzu knife man said on TV, there's more. By Parseval's theorem they are square-summable, too!