Interesting Integral

I found this result while studying indefinite integral calculus last year.

\(\displaystyle I_{n}\,=\, \int e^{t} \cdot \left[f(t)+(-1)^{n-1}\cdot f^{(n)}(t)\right]dt\,=\,e^{t}\cdot\left[f(t)-f^{\prime}(t)+f^{\prime \prime}(t)-...+(-1)^{n-1}f^{(n-1)}(t)\right]+c\), where \(f^{(k)}(x)\) denotes the \(k\)th derivate of \(f(x)\) and \(n \,\in\, \mathbb{Z^{+}} \).

This is basically the generalization of the integral \(\displaystyle \int e^{t} \cdot \left[f(t)+f^{\prime}(t)\right] dt\,=\,e^{t}\cdot f(t)\,+\,c\).

\(\huge \text{Proof :-}\)

It can be seen that \( \displaystyle I_{1}\,=\, \int e^{t} \cdot \left[f(t)+f^{\prime}(t)\right] dt \). Using integration by parts,

\(\displaystyle I_{1}\,=\, \int e^{t} \cdot \left[f(t)+f^{\prime}(t)\right] dt\,=\,\int \underbrace{e^{t}}_{\text{II}}\cdot \underbrace{f(t)}_{\text{I}} \,dt + \int e^{t} \cdot f^{\prime}(t) \,dt \,=\,e^{t}\cdot f(t) \,-\, \int e^{t} \cdot f^{\prime}(t)\,dt+ \int e^{t} \cdot f^{\prime}(t)\,dt\,=\,e^{t}\cdot f(t) + c \ \).

So, \(I_{1}\) is true. Assume, \(I_{r}\) is true i.e \(\displaystyle I_{r}\,=\, \int e^{t} \cdot \left[f(t)+(-1)^{r-1}\cdot f^{(r)}(t)\right]dt\,=\,e^{t}\cdot\left[f(t)-f^{\prime}(t)+f^{\prime \prime}(t)-...+(-1)^{r-1}f^{(r-1)}(t)\right]+c\).


\[ I_{r+1}\,=\,\int e^{t} \cdot \left[f(t)\,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\,=\,\int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{blue}(-1)^{r-1} \cdot f^{(r)}(t)}\,-\,\mathbin{\color{blue}(-1)^{r-1} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\]

\[\implies\int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{blue}(-1)^{r-1} \cdot f^{(r)}(t)}\,+\,\mathbin{\color{blue}(-1)^{r} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\,=\, \int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{blue}(-1)^{r-1} \cdot f^{(r)}(t)}\right]dt \,+\, \int e^{t} \cdot \left[\mathbin{\color{blue}(-1)^{r} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\]

\[\implies I_{r}\,+\, (-1)^{r} \cdot \int \left[\underbrace{e^{t}}_{\text{II}}\cdot \underbrace{\mathbin{\color{blue} f^{(r)}(t)}}_{\text{I}} \,+\,e^{t}\cdot f^{(r+1)}(t)\right]\,dt\,=\,I_{r}+ (-1)^{r} \cdot \left[e^{t} \cdot f^{(r)}(t)\,-\,\int e^{t} \cdot f^{(r+1)}(t)\,dt\,+\,\int e^{t} \cdot f^{(r+1)}(t)\,dt\right]\,=\,I_{r}+(-1)^{r} \cdot e^{t}\cdot f^{(r)}(t)\].

From our assumption, it follows that,

\[I_{r+1}\,=\,e^{t}\cdot\left[f(t)-f^{\prime}(t)+f^{\prime \prime}(t)-...+(-1)^{r}f^{(r-1)}(t)+(-1)^{r}\cdot f^{(r)}(t)\right]+c\].

So, \(I_{r+1}\) is true. Hence, \(I_{r}\) is true \(\implies\) \(I_{r+1}\) is true. So, by induction, \(I_{n}\) is true \(\forall \,n \,\in\,\,\mathbb{Z^{+}}\).

Also, integrals of the form \(\displaystyle \int \left[f(\log x)+(-1)^{n-1}\cdot f^{(n)}(\log x)\right]\,dx\) can be reduced to the aforementioned integral by the substitution \(t=\log (x)\).

Note by Aditya Sky
10 months, 3 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)


Sort by:

Top Newest

Awesome proof - dont have time to check its validity but quick glance looks correct!

Elethelectric Penguin - 9 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...