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# 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,

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

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

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

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 $$t=\log (x)$$.

2 months, 2 weeks ago

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Awesome proof - dont have time to check its validity but quick glance looks correct! · 4 weeks ago

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