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

Question

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

Answer 1

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 $\frac{\varepsilon}{2}$.

Answer 2

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!