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There was a post asking for help defining $\int x^{2} e^{x} \, \mathit{dx}$ using Integration by Parts. (See question 2123271.) A comment was made that the integral could be defined without using Integration by Parts. "If $f(t) = \int e^{tx} \, \mathit{dx}$, $f^{\prime\prime}(1) = \int x^{2} e^{x} \, \mathit{dx}$."

What is the definition of $f^{\prime}(t)$ and $f^{\prime\prime}(t)$?

The comment on the post states that the value of $f^{\prime\prime}(t)$ at $1$ helps to show that \begin{equation*} \int x^{2} e^{x} \, \mathit{dx} = (x^{2} - 2x + 2)e^{x} . \end{equation*} Please explain.

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The trick is that we know how to compute this integral exactly:

$$ \int e^{tx}dx $$ The result is a function of $t$ (including the constant of integration)

$$ \int e^{tx} dx = \frac{1}{t}e^{tx} + C_1(t) $$ So, this means (assuming everything is mathematically "kosher", which here it is),

$$ \frac{d}{dt}\int e^{tx}dx = \frac{d}{dt} \left(\frac{1}{t}e^{tx} + C\right) = -\frac{1}{t^2}e^{tx} + \frac{x}{t}e^{tx} + C_2(t) $$ But on the other hand, (again, assuming we can "pull the derivative inside the integral", which we can here):

$$ \frac{d}{dt}\int e^{tx}dx = \int\frac{\partial}{\partial t}e^{tx} dx = \int xe^{tx}dx $$ So we have arrived at the nice formula

$$ \int xe^{tx} dx = \left(\frac{x}{t} - \frac{1}{t^2}\right)e^{tx} + C_2(t) $$ You can evaluate this at any value of $t$; $t=1$ will give $\int xe^xdx$. If you do this trick again (take a second derivative with respect to $t$) and evaluate the result at $t=1$, you can arrive at the formula for $\int x^2e^{x}dx$. Note that the "constant of integration" $C_1(t)$ can be chosen to be any function of $t$, so after evaluating at some $t$ value, we still have an "arbitrary constant" - if you were trying to evaluate a definite integral by this method, you could choose $C_1(t)\equiv 0$.

icurays1
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Let $f(t)=\int e^{tx}\,dx=\frac{e^{tx}}{t}+C$. Taking the derivative, $f'(t)$, of $f(t)$ with respect to $t$ reveals

$$f'(t)=\int xe^{tx}\,dx=x\frac{e^{tx}}{t}-\frac{e^{tx}}{t^2}+C \tag 1$$

Note that at $t=1$, $(1)$ becomes $f'(1)=\int xe^x\,dx=(x-1)e^x+C$.

Differentiating $(1)$ with respect to $t$ yields

$$f''(t)=\int x^2e^{tx}\,dx=x^2\frac{e^{tx}}{t}-2x\frac{e^{tx}}{t^2}+2\frac{e^{tx}}{t^3}+C\tag 2$$

Evaluating $(2)$ at $t=1$, we obtain

$$f''(1)=\int x^2 e^{x}\,dx=(x^2-2x+2)e^x+C$$

as was to be shown!

Mark Viola
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  • According to the Fundamental Theorem of Calculus, $f^{\prime}(t) = \frac{\mathit{d}}{\mathit{dt}} \int e^{tx} , \mathit{dx} = \int \frac{\mathit{d}}{\mathit{dt}} , e^{tx} , \mathit{dx} = \int xe^{tx} , \mathit{dx}$. –  Apr 07 '17 at 18:24
  • Is that correct? –  Apr 07 '17 at 18:25
  • @AgalnamedDesire it's not the fundamental theorem of calculus, it's called the Leibniz Integral Rule, or "differentiation under the integral sign" – icurays1 Apr 07 '17 at 18:28
  • The FTOC states that if $F(x)=\int_a^x f(t),dt$, then $F'(x)=f(x)$ if $f$ is continuous. This is differentiating under the integral. – Mark Viola Apr 07 '17 at 18:35