Alternative to Integration by Parts?

Click For Summary
SUMMARY

This discussion explores alternatives to the integration by parts technique, specifically in the context of integrating the product of two functions. The Tabular Method is highlighted as a more organized approach for multiple integrations by parts, particularly for integrals like \int{x^5}{e^x}dx. Additionally, the use of differentiation with respect to a parameter and Taylor expansions is presented as a more efficient method for complex integrals, allowing for the extraction of derivatives without repetitive integration by parts. The discussion concludes that while integration by parts is effective, there are indeed simpler and more intuitive methods available.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts
  • Familiarity with the Tabular Method for integration
  • Knowledge of Taylor series and their applications in calculus
  • Basic concepts of differentiation with respect to parameters
NEXT STEPS
  • Research the Tabular Method for integration in depth
  • Learn about differentiation with respect to parameters in calculus
  • Explore Taylor series expansions and their applications in integration
  • Investigate advanced integration techniques beyond traditional methods
USEFUL FOR

Mathematicians, calculus students, educators, and anyone interested in exploring alternative integration techniques and enhancing their problem-solving skills in calculus.

Masna
Messages
4
Reaction score
0
Hey all!

I was recently refreshing my memory of integration by parts via some personal reading when I thought, there must be a better way. Integration by parts (while creative in that it integrates the entire product rule) feels very arbitrary to what it's attempting to calculate (at least concerning the integral of a product of two functions).

So, I ask, are there any alternatives to integration by parts? Perhaps, any less messy methods? Don't take this the wrong way: integration by parts is brilliant. But I still feel that there has to be a simpler, much more intuitive way to calculate the integral of the product of two functions.

If there aren't any defined alternatives, is there any work being done by any mathematicians (publicly) on a better way to complete said task?

I personally spent a few hours trying to make some connections that might lead me finding an alternative method, I didn't find an alternative method.

Thanks!
 
Physics news on Phys.org
I tend to use the Tabular Method, which is just a "rearranged" way of integration by parts and is a little nicer if you have some type of integral that requires multiple stages of integration by parts, e.g...

\int{x^5}{e^x}dx

For a more exotic integral that you won't necessarily notice it requires integration by parts, I will revert back to the original way.

Other than the tabular method, I have not seen many more techniques, other than using tables...
 
Differentiation w.r.t. a suitably chosen parameter (or Taylor expanasions w.r.t. that parameter) is often far more efficent than partial integration.

Take e.g. the example given by pbandjay above:

\int x^{5}\exp(x)dx

The smart way to compute this is by replacing x^{5} by \exp(px). Then, the fifth derivative w.r.t. p at p = 0 will yield the answer. However, computing that fifth derivative will aslo be a bit messy, albeit far less messy that doing a partial integration 5 times. Now, computing high order derivatives is more easily done by obtaining the Taylor expansion and then extracting the derivatives from that.

So, the method works as follows. We have:

\int \exp(px)\exp(x)dx = \int \exp\left[(p+1)x\right]dx=\frac{\exp\left[(p+1)x\right]}{p+1}=\exp(x)\frac{\exp(px)}{1+p}

If you expand the factor

\frac{\exp(px)}{1+p}

in powers of p by multiplying the Taylor series of thegeometric series and of the exponential function and extract the coefficient of p^5, you easily find that it is given by:

\frac{x^5}{5!}-\frac{x^4}{4!}+\frac{x^3}{3!}-\frac{x^2}{2!}+x-1

The coeficient of p^5 of the integral

\int \exp(px)\exp(x)dx

is 1/5! times the desired integral, so we find:

\int x^{5}\exp(x)dx = \left(x^5-5x^4+20x^3-60x^2+120x-120\right)\exp(x) + c
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K