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Can $\int xe^{-x} dx$ ever be solved by integration by substitution without using parts. Or does, as I suspect, substitution fail to yield a solution in this case.

Seems that we can't get a reciprocal to cancel $x$ out under any circumstance. I know how to solve with parts, please don't use this.

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  • $\begingroup$ If it does fail, mathematically, why? $\endgroup$ – Timothy Mar 4 '16 at 11:05
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    $\begingroup$ Sorry not to answer your question but it made wondering why don't you want to use integration by parts? Using it would not be a case of "killing a fly with a rocket-launcher" $\endgroup$ – Taladris Mar 4 '16 at 11:05
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    $\begingroup$ Consider $-\int e^{-\alpha x}=\alpha^{-1}e^{-\alpha x}$. Differentiate both side w.r.t. $\alpha$ and set $\alpha=1$. $\endgroup$ – A.S. Mar 4 '16 at 11:07
  • $\begingroup$ @A.S. I haven't done calculus in 3 years, so could you help me out slightly more? How do I get from $\int xe^{-x} dx$ to $-\int e^{-\alpha x}$? Where does the leading $x$ go? $\endgroup$ – Timothy Mar 4 '16 at 11:13
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    $\begingroup$ PLEASE DO NOT SHOUT! IT IS VERY ANNOYING TO READ TITLES WRITTEN LIKE THAT! THANKS! $\endgroup$ – Najib Idrissi Mar 4 '16 at 12:16
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Differentiation under the integral sign.

The integral is precisely $$-\frac{1}{a} \dfrac{d}{da} \int e^{-a x} \ dx$$ evaluated at $a=1$.

But that is $$-\dfrac{1}{a} \dfrac{d}{da} \left(-\frac{1}{a} e^{-ax} \right) = -\frac{1}{a} \left( \frac{e^{-ax}}{a^2} + \frac{e^{-a x}x}{a} \right)$$

Evaluating at $1$ yields $$-(e^{-x} + x e^{-x}) = -e^{-x}(1+x)$$

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By method of undetermined coefficients

Guessing the anti-derivative $F(x)$ is in the form of $(A+Bx)e^{-x}$, then $F'(x)=-Ae^{-x}+Be^{-x}-Bxe^{-x}$ implying $B=-1$ and $B-A=0$.

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Consider a basis $\mathcal{B} = \{e^{-x}, xe^{-x}\}$. Differentiating each element of the basis we have

\begin{align*} \frac{d}{dx}(e^{-x}) &= -e^{-x}\\ \frac{d}{dx}(xe^{-x}) &=e^{-x} + -xe^{-x} \end{align*}

The matrix representation of the derivative operator with respect to $\mathcal{B}$ is

$$T = \begin{bmatrix} -1 & 1\\ 0 & -1 \end{bmatrix}$$

and

$$T^{-1} = \begin{bmatrix} -1 -1\\ 0 -1 \end{bmatrix}.$$

So

$$\int xe^{-x} \operatorname{d}\!x = T^{-1}\begin{bmatrix} 0\\ 1 \end{bmatrix} = -e^{-x} - xe^{-x}.$$

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  • $\begingroup$ i like it +1, where have you fond this method please @jessicaK ? $\endgroup$ – Hamza Mar 4 '16 at 12:34
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    $\begingroup$ @Hamza I'm sorry, I don't have a source for you. I've never seen it in a book, probably because it isn't very useful. I found this one day after first learning about half-derivatives and trying to figure out how one might define a half-integral; the square root of a matrix seemed like the natural course of action. $\endgroup$ – JessicaK Mar 4 '16 at 12:46
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Since a solution by substitution is required, substitute $y=x$, then proceed as in one of the other answers that involve neither parts, nor substitution.

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    $\begingroup$ This is… facetious. $\endgroup$ – Patrick Stevens Mar 4 '16 at 11:25
  • $\begingroup$ How is this helpful to me? Since a solution to the Riemann hypothesis is required. State the Riemann hypothesis and proceed to complete a correct proof. $\endgroup$ – Timothy Mar 4 '16 at 12:26
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Let's write $$e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}$$ then \begin{align} \int x e^{-x}dx&=\int \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n!}dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\int x^{n+1}dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+2}}{n!(n+2)}+k\\ &=\sum_{n=0}^{\infty}\frac{(n+1)(-1)^{n+2}x^{n+2}}{(n+2)!}+k\\ &=-\sum_{n=0}^{\infty}\frac{n(-1)^{n}x^{n+1}}{(n+1)!}+k\\ &=-\sum_{n=0}^{\infty}\big(\frac{1}{n!}-\frac{1}{(n+1)!}\big)(-1)^{n}x^{n+1}+k\\ &=-\sum_{n=0}^{\infty}\big(\frac{1}{n!}\big)(-1)^{n}x^{n+1}+\sum_{n=0}^{\infty}\big(\frac{1}{(n+1)!}\big)(-1)^{n}x^{n+1}+k\\ &=-xe^{-x}-e^{-x}+1+k\\ &=-xe^{-x}-e^{-x}+k'\\ \end{align}

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