$\lim_{n \to \infty} \int_{0}^{1} e^{x^2} \sin(nx) dx$ $$\lim_{n \to \infty} \int_{0}^{1} e^{x^2} \sin(nx) dx$$
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
 A: 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!
A: 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 $\frac{\varepsilon}{2}$.
A: If you are familiar with the basic of Fourier series, one  can approach this though either the Riemann-Lebesgue lemma, to by Féjer's formula.
For example, since $\mathbb{1}_{[0,1]}(x)e^{x^2}$  and $x\mapsto\sin x$ is $2\pi$-periodic,  Féjer's formula gives
$$
\lim_n\int \mathbb{1}_{[0,1]}(x)e^{x^2}\sin(nx)\,dx=\Big(\frac{1}{2\pi}\int^{2\pi}_0\sin(x)\,dx\Big)\int\mathbb{1}_{[0,1]}(x)e^{x^2}\,dx=0$$
