# Differentiating under integral sign — trig counterexample

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?

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The last term should have a negative sign, but other than that I don't see what's wrong. This is not a good application of differentiation under the integral sign, though, since generally people only care about integer $n$; I would instead compute $\int_{-\pi}^{\pi} e^{tx} \cos nx \, dx$. –  Qiaochu Yuan Jul 20 '11 at 15:53
Your Laplace transform example is complicated. Instead, I should just evaluate the RHS for arbitrary n, even though I just want integer n. This was for my multivariate calculus class, which I am teaching :-/ –  john mangual Jul 20 '11 at 16:09

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

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

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• If $u$ and $v$ are functions of $x$ and dashes denote differentiation and suffixes integration with respect to $x$, then $$\small\int uv \ dx = uv_{1} -u'v_{2}+ u''v_{3} - u'''v_{4} + \cdots + (-1)^{n-1}u^{(n-1)}v_{n} + (-1)^{n} \int u^{(n)}\cdot v_{n} \ dx$$
What do the $v_n$ stand for in your formula? –  Pedro Tamaroff Feb 26 '12 at 17:43
I read you answer. You clarify $v_n$ stands for integration. –  Pedro Tamaroff Feb 27 '12 at 18:03