Differentiating under integral sign — trig counterexample

Question

I really hate integration by parts, so when faced with $\int_{-\pi}^\pi x^2 \cos n x \, dx$ I tried writing it as $$\int_{-\pi}^\pi x^2 \cos n x \, dx =
\frac{d}{dn} \int_{-\pi}^\pi x \sin n x \, dx = \frac{d^2}{dn^2} \int_{-\pi}^\pi \cos n x \, dx $$1

I have done something wrong. The integrand is continuously differentiable with respect to n and I thought that was enough. How can I get differentiating under the integral sign to work?

Answer 1

It did work (except you are missing a negative sign). Remember $n$ is not always an integer, so that $$-\int_{-\pi}^\pi \cos(nx)dx=-\frac{2\sin(\pi n)}{n}.$$

Then $$\frac{d^2}{dn^2} \left(-\int_{-\pi}^\pi \cos(nx)dx\right) =\frac{d}{dn} \left( -\frac{2\pi\cos (\pi n)}{n}+\frac{2\sin(\pi n)}{n^2}\right)$$

$$=\frac{2\pi^2\sin(\pi n)}{n}-\frac{4\sin(\pi n)}{n^3}+\frac{4\pi\cos(\pi n)}{n^2}.$$

Answer 2

Well, if there are cases in which you can't avoid integration by parts, then please use the below formula:

• If $u$ and $v$ are functions of $x$ and dashes denote differentiation and suffixes integration with respect to $x$, then $$\small\int uv \ dx = uv_{1} -u'v_{2}+ u''v_{3} - u'''v_{4} + \cdots + (-1)^{n-1}u^{(n-1)}v_{n} + (-1)^{n} \int u^{(n)}\cdot v_{n} \ dx$$