MHB Solving Limits of Integrals: Advice & Tips

[Vali]

View attachment 8746

I tried to use integration by parts.
I took f(x)=arctan(x) => f'(x)= 1/x^2+1
g'(x)=cos(nx) => g(x)= sin(nx)/n
So I get sin(nx)/n * arctan(x) - integral from 0 to 1 from sin(nx)/n(x^2+1)
How to continue ?
I'm always getting stuck with this kind of exercises ( limits of integrals ) because I don't know how to replace n (infinity) in functions and I noticed that I have to use intervals and inequalities to resolve this kind of limits.
Some ideas?

 

[MarkFL]

I've moved this thread here to our "Calculus" forum since it involves integration.

I would begin by writing:

$$L=\lim_{n\to\infty}\left(\int_0^1 \arctan(x)\cos(nx)\,dx\right)$$

Using IBP in the integral, let's try:

$$u=\arctan(x)\implies du=\frac{1}{x^2+1}\,dx$$

$$dv=\cos(nx)\,dx\implies v=\frac{1}{n}\sin(nx)$$

And so:

$$\int_0^1 \arctan(x)\cos(nx)\,dx=\left[\frac{\arctan(x)\sin(nx)}{n}\right]_0^1-\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx=\frac{\pi\sin(n)}{4n}-\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx$$

Hence:

$$L=-\lim_{n\to\infty}\left(\frac{1}{n}\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx\right)$$

Now, it seems to me considering an increasingly oscillating but bounded sinusoid (which varies from -1 to 1) we can state:

$$\int_0^1 \frac{\sin(nx)}{x^2+1}\,dx\le 2\int_0^1 \frac{1}{x^2+1}\,dx=\frac{\pi}{2}$$

And so:

$$L=-\lim_{n\to\infty}\left(\frac{\pi}{2n}\right)=0$$

 

[MountEvariste]

Riemann-Lebesgue lemma:

If $f$ is a real or complex valued measurable function defined on $I$ and Lebesgue integrable over $I$ then:

$\displaystyle \lim_{|\lambda| \to \infty}\int_{I} f(x) \cos({\lambda x})\,\mathrm{d}{x} = \lim_{|\lambda| \to \infty} \int_{I} f(x) \sin({\lambda x})\,\mathrm{d}{x}=0.$​