[Math] Generalising integration by parts for the product of more than two functions

integration

Just as the product rule can be generalised to the product of more than two functions, i.e. $$\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ]
= \sum_{i=1}^k \left(\frac{d}{dx} f_i(x) \prod_{j\ne i} f_j(x) \right)
= \left( \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right),$$ is there a way to generalise integration by parts to evaluate $$\int f_1(x) f_2(x) \cdots f_n(x) dx \qquad ?$$

For contextual purposes only, I'm trying to evaluate by hand

$$\int x \cosh(x+1)e^x\sin(x) dx \quad
.
$$

I know that I could let $u=x \cosh(x+1)$ and $v \ '=e^x\sin(x)$ but that would require integration by parts to be performed at least three times.

Is there a more-efficient way to evaluate this integral using the proposed 'generalisation'?

Thanks!

Best Answer

The Wikipedia article on integration by parts gives you the generalization you're looking for:

$$\Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x),$$

where $u_i(x)$ are your $n$ functions of $x$ that are terms of the product that comprise your integrand.

Related Solutions

Integration – Constant of Integration During Integration by Parts

Here's one way to think about it: we reach this step: $$ \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx $$ And for our next step, we'd like to add $\int e^x \sin x \, dx$ on both sides, so we do. $$ \int e^x \sin x \, dx + \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx + \int e^x \sin x \, dx \\ \int [e^x \sin x + e^x \sin x] \, dx = e^x \sin x - e^x \cos x + \int [e^x \sin x - e^x \sin x] \, dx \\ \int 2[e^x \sin x] \, dx = e^x \sin x - e^x \cos x + \int 0\, dx \\ 2\int e^x \sin x \, dx = e^x \sin x - e^x \cos x + C $$ The intuition is that when we take the antiderivative twice in the same equation, we don't guarantee that the result is the same each time. So, the equation $$ \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx $$ Is really something to the effect of $$ F(x) + C_1 = e^x \sin x - e^x \cos x - [F(x) + C_2] $$ Where $F$ is some antiderivative of $e^x \sin x$.

Integration – Advanced Techniques for Definite and Indefinite Integrals

Here are a few. The first one is included because it's not very well known and is not general, though the ones that follow are very general and very useful.

A great but not very well known way to find the primitive of $f^{-1}$ in terms of the primitive of $f$, $F$, is (very easy to prove: just differentiate both sides and use the chain rule): $$ \int f^{-1}(x)\, dx = x \cdot f^{-1}(x)-(F \circ f^{-1})(x)+C. $$

Examples:

$$ \begin{aligned} \displaystyle \int \arcsin(x)\, dx &= x \cdot \arcsin(x)- (-\cos\circ \arcsin)(x)+C \\ &=x \cdot \arcsin(x)+\sqrt{1-x^2}+C. \end{aligned} $$

$$ \begin{aligned} \int \log(x)\, dx &= x \cdot \log(x)-(\exp \circ \log)(x) + C \\ &= x \cdot \left( \log(x)-1 \right) + C. \end{aligned} $$

This one is more well known, and extremely powerful, it's called differentiating under the integral sign. It requires ingenuity most of the time to know when to apply, and how to apply it, but that only makes it more interesting. The technique uses the simple fact that $$ \frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y. $$

Example:

We want to calculate the integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$. To do that, we unintuitively consider the more complicated integral $\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$ instead.

Let $$ I(t)=\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx,$$ then $$ I'(t)=-\int_{0}^{\infty} e^{-tx} \sin(x) dx=\frac{e^{-t x} (t \sin (x)+\cos (x))}{t^2+1}\bigg|_0^{\infty}=\frac{-1}{1+t^2}.$$

Since both $I(t)$ and $-\arctan(t)$ are primitives of $\frac{-1}{1+t^2}$, they must differ only by a constant, so that $I(t)+\arctan(t)=C$. Let $t\to \infty$, then $I(t) \to 0$ and $-\arctan(t) \to -\pi/2$, and hence $C=\pi/2$, and $I(t)=\frac{\pi}{2}-\arctan(t)$.

Finally, $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0) = \frac{\pi}{2}-\arctan(0) = \boxed{\frac{\pi}{2}}. $$

This one is probably the most commonly used "advanced integration technique", and for good reasons. It's referred to as the "residue theorem" and it states that if $\gamma$ is a counterclockwise simple closed curve, then $\displaystyle \int_\gamma f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( f, a_k )$ . It will be difficult for you to understand this one without knowledge in complex analysis, but you can get the gist of it with the wiki article. Example:

We want to compute $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$. The poles of our function $f(z)=\frac{x^2}{1+x^4}$ in the upper half plane are $a_1=e^{i \frac{\pi}{4}}$ and $a_2=e^{i \frac{3\pi}{4}}$. The residues of our function at those points are $$\operatorname{Res}(f,a_1)=\lim_{z\to a_1} (z-a_1)f(z)=\frac{e^{i \frac{-\pi}{4}}}{4},$$ and $$\operatorname{Res}(f,a_2)=\lim_{z\to a_2} (z-a_2)f(z)=\frac{e^{i \frac{-3\pi}{4}}}{4}.$$ Let $\gamma$ be the closed path around the boundary of the semicircle of radius $R>1$ on the upper half plane, traversed in the counter-clockwise direction. Then the residue theorem gives us ${1 \over 2\pi i} \int_\gamma f(z)\,dz=\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_2)={1 \over 4}\left({1-i \over \sqrt{2}}+{-1-i \over \sqrt{2}}\right)={-i \over 2 \sqrt{2}}$ and $ \int_\gamma f(z)\,dz= {\pi \over \sqrt{2}}$. Now, by the definition of $\gamma$, we have: $$\int_\gamma f(z)\,dz = \int_{-R}^R \frac{x^2}{1+x^4} dx + \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz = {\pi \over \sqrt{2}}.$$ For the integral on the semicircle $$ \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz, $$ we have $$ \begin{aligned} \left| \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \right| &\leq \int_0^\pi \left| {i (R e^{it})^3 \over 1+(R e^{it})^4} \right| dz \\ &\leq \int_0^\pi {R^3 \over R^4-1} dz={\pi R^3 \over R^4-1}. \end{aligned} $$ Hence, as $R\to \infty$, we have ${\pi R^3 \over R^4-1} \to 0$, and hence $\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \to 0$. Finally, $$ \begin{aligned} \int_{-\infty}^\infty \frac{x^2}{1+x^4} dx &= \lim_{R\to \infty} \int_{-R}^R \frac{x^2}{1+x^4} dx \\ &= \lim_{R\to \infty} {\pi \over \sqrt{2}}-\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz =\boxed{{\pi \over \sqrt{2}}}. \end{aligned} $$

My final "technique" is the use of the mean value property for complex analytic functions, or Cauchy's integral formula in other words: $$ \begin{aligned} f(a) &= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-a}\, dz \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} f\left(a+e^{ix}\right) dx. \end{aligned} $$