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

Consider,

$I_{r+1}\,=\,\int e^{t} \cdot \left[f(t)\,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\,=\,\int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{#3D99F6}(-1)^{r-1} \cdot f^{(r)}(t)}\,-\,\mathbin{\color{#3D99F6}(-1)^{r-1} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt$

$\implies\int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{#3D99F6}(-1)^{r-1} \cdot f^{(r)}(t)}\,+\,\mathbin{\color{#3D99F6}(-1)^{r} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt\,=\, \int e^{t} \cdot \left[f(t)\,+\,\mathbin{\color{#3D99F6}(-1)^{r-1} \cdot f^{(r)}(t)}\right]dt \,+\, \int e^{t} \cdot \left[\mathbin{\color{#3D99F6}(-1)^{r} \cdot f^{(r)}(t)} \,+\,(-1)^{r} \cdot f^{(r+1)}(t)\right]\,dt$

$\implies I_{r}\,+\, (-1)^{r} \cdot \int \left[\underbrace{e^{t}}_{\text{II}}\cdot \underbrace{\mathbin{\color{#3D99F6} f^{(r)}(t)}}_{\text{I}} \,+\,e^{t}\cdot f^{(r+1)}(t)\right]\,dt\,=\,I_{r}+ (-1)^{r} \cdot \left[e^{t} \cdot f^{(r)}(t)\,-\,\int e^{t} \cdot f^{(r+1)}(t)\,dt\,+\,\int e^{t} \cdot f^{(r+1)}(t)\,dt\right]\,=\,I_{r}+(-1)^{r} \cdot e^{t}\cdot f^{(r)}(t)$.

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 $t=\log (x)$.