Integrating $\int_0^\pi x^4\cos(nx)\,dx$ with Feynman trick?

I should make this integral : $$\displaystyle\int_0^\pi x^4\cos(nx)\,dx$$

Usually you would integrate 4 times by part.

I was wondering if there would be a more direct way, something like using the Leibniz rule (aka Feynman trick) ?

• There is tabular integration, which is just a fast way of doing integration by parts. It's also known as the "Stand and Deliver" method, because it's featured in that movie. – Adrian Keister Sep 27 '19 at 13:53

If we consider the following integral, with $$z\in R$$: $$\int_0^\pi \cos(zx)dx=\frac{\sin(\pi z)}{z}$$ Then all we need to do is to take $$4$$ derivatives with respect to $$z$$ (on both sides) then set $$z=n$$ to get the integral in the question, since: $$\frac{d}{dz}\left(\frac{\sin(\pi z)}{z}\right)=-\int_0^\pi x\sin(zx)dx$$ $$\frac{d^2}{dz^2}\left(\frac{\sin(\pi z)}{z}\right)=-\int_0^\pi x^2\cos(zx)dx$$ $$\frac{d^3}{dz^3}\left(\frac{\sin(\pi z)}{z}\right)=\int_0^\pi x^3\sin(zx)dx$$ $$\frac{d^4}{dz^4}\left(\frac{\sin(\pi z)}{z}\right)=\int_0^\pi x^4\cos(zx)dx$$

Of course in case you're looking for $$\int_0^\pi x^3 \cos(nx)dx$$ then you might want to consider initially $$\int_0^\pi \sin(zx)dx$$ and proceed as above.

I would also like to mention that this method also works for other integrals, for example let's take: $$\int_0^1 x^9 \ln^5 xdx$$ All there is needed to do is to consider: $$\int_0^1 x^z dx=\frac{1}{z+1}\Rightarrow \int_0^1 x^z \ln xdx=\frac{d}{dz}\left(\frac{1}{z+1}\right)$$ $$\Rightarrow \int_0^1 x^9 \ln^5 x dx= \lim_{z\to 9}\frac{d^5}{dz^5}\left(\frac{1}{z+1}\right)$$

• Wrong Integration variable, sent correction – eyal karni Jan 23 at 4:15

Define $$f(a) = \int_0^\pi e^{ax} dx$$ Use Leibniz rule to differentiate with respect to $$a$$, $$4$$ times, set $$a=ni$$ then take the real part.

EDIT: Sorry, I didn't read the question fully; this doesn't use the Leibniz rule.

If I recall, the cumulative distribution function for the Erlang distribution works for complex parameters. Your integral is $$\Re\left(\int_0^\pi x^4 e^{inx}dx\right) = \Re\left(\frac{\Gamma(5)}{(-in)^5} \int_0^\pi \frac{(-in)^5 x^4 e^{inx}}{\Gamma(5)}dx\right)$$ where $$\Re$$ denotes the real part. The integral is the cumulative distribution function evaluated at $$\pi$$, namely $$\int_0^\pi \frac{(-in)^5 x^4 e^{inx}}{\Gamma(5)}dx = 1 - \sum_{j=0}^4 \frac{(-in\pi)^j}{j!}e^{in\pi},$$ i.e. $$\int_0^\pi x^4 \cos(nx)dx = \Re\left(\frac{24i}{n^5}\left(1 - \sum_{j=0}^4 \frac{(-in\pi)^j}{j!}e^{in\pi}\right)\right).$$

Take $$\int$$ as operator D^(-1) (1/D)e^(ax)x^4=e^(ax)(1/(D+a))x^4=e^(ax)(1/a)(1+D/a)^-1 {x^4} =(1/a)e^(ax){1-D/a+D^2 /a^2 +D^3/a^3-d^4/a^4}x^4

We can take a as $$in$$ and proceed to obtain the real part of the expression. [$$\begin{gathered} \int {{e^{ax}}} {x^n}dx = \frac{1}{D}\left\{ {{e^{ax}}.{x^4}} \right\} = {e^{ax}}\frac{1}{{D + a}}\left\{ {{x^4}} \right\} \hfill \\ = \frac{{{e^{ax}}}}{a}\frac{1}{{\left( {1 + \frac{D}{a}} \right)}}\left\{ {x4} \right\} = \frac{{{e^{ax}}}}{a}{\left( {1 + \frac{D}{a}} \right)^{ - 1}}\left\{ {{x^4}} \right\} \hfill \\ = \frac{{{e^{ax}}}}{a}\left( {1 - \left( {\frac{D}{a}} \right) + {{\left( {\frac{D}{a}} \right)}^2} - {{\left( {\frac{D}{a}} \right)}^3} + {{\left( {\frac{D}{a}} \right)}^4} - \cdots } \right)\left\{ {{x^4}} \right\} \hfill \\ = \frac{{{e^{ax}}}}{a}\left\{ {{x^4} - \frac{{4{x^3}}}{a} + \frac{{4.3{x^2}}}{{{a^2}}} - \frac{{4.3.2x}}{{{a^3}}} + \frac{{4.3.2.1}}{{{a^4}}}} \right\} \hfill \\ Let\,a = j.n;and\,obtain\,the\,real\,part\,of\,the\,result. \hfill \\ \end{gathered}$$ ]