Interesting Integral

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

In=et[f(t)+(1)n1f(n)(t)]dt=et[f(t)f(t)+f(t)...+(1)n1f(n1)(t)]+c\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)f^{(k)}(x) denotes the kkth derivate of f(x)f(x) and nZ+n \,\in\, \mathbb{Z^{+}} .

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

Proof :-\huge \text{Proof :-}

It can be seen that I1=et[f(t)+f(t)]dt \displaystyle I_{1}\,=\, \int e^{t} \cdot \left[f(t)+f^{\prime}(t)\right] dt . Using integration by parts,

I1=et[f(t)+f(t)]dt=etIIf(t)Idt+etf(t)dt=etf(t)etf(t)dt+etf(t)dt=etf(t)+c \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, I1I_{1} is true. Assume, IrI_{r} is true i.e Ir=et[f(t)+(1)r1f(r)(t)]dt=et[f(t)f(t)+f(t)...+(1)r1f(r1)(t)]+c\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.


Ir+1=et[f(t)+(1)rf(r+1)(t)]dt=et[f(t)+(1)r1f(r)(t)(1)r1f(r)(t)+(1)rf(r+1)(t)]dt 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

    et[f(t)+(1)r1f(r)(t)+(1)rf(r)(t)+(1)rf(r+1)(t)]dt=et[f(t)+(1)r1f(r)(t)]dt+et[(1)rf(r)(t)+(1)rf(r+1)(t)]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

    Ir+(1)r[etIIf(r)(t)I+etf(r+1)(t)]dt=Ir+(1)r[etf(r)(t)etf(r+1)(t)dt+etf(r+1)(t)dt]=Ir+(1)retf(r)(t)\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,

Ir+1=et[f(t)f(t)+f(t)...+(1)rf(r1)(t)+(1)rf(r)(t)]+cI_{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, Ir+1I_{r+1} is true. Hence, IrI_{r} is true     \implies Ir+1I_{r+1} is true. So, by induction, InI_{n} is true nZ+\forall \,n \,\in\,\,\mathbb{Z^{+}}.

Also, integrals of the form [f(logx)+(1)n1f(n)(logx)]dx\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)t=\log (x).

Note by Aditya Sky
4 years, 2 months ago

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

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ground braking

Nahom Assefa - 3 years ago

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