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I saw from Simply Beautiful Art that the indefinite integral $\int e^{tx} \, dt$ may be defined by a technique that is not integration by parts. I tried to find his comment that he left on a post this morning. Can someone help me find it?

  • @Simply Beautiful Art Can you give me the link to the post? –  Feb 01 '17 at 19:01
  • If you are integrating with respect to $t$, and that is what it looks like according to the differential, then of course no integration by parts is needed. – imranfat Feb 01 '17 at 19:04
  • I think you are talking about differentiation under the integral sign ? – Vivek Kaushik Feb 01 '17 at 19:06
  • @imranfat Yes, you are right. I do not remember the post to which Simply Beautiful Art replied. I looked at the posts today but didn't find it. –  Feb 01 '17 at 19:24
  • @imranfat Here is the link. http://math.stackexchange.com/questions/2123271/methods-for-choosing-u-and-dv-when-integrating-by-parts –  Feb 01 '17 at 22:09

3 Answers3

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I mean I'm not sure about which article you read, but you could use the substitution: $u = tx$, $\partial u = x\partial t$, to integrate it.

Felicio Grande
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$$\int { { e }^{ tx }dt } =\frac { 1 }{ x } \int { { e }^{ tx }d\left( xt \right) } =\frac { { e }^{ tx } }{ x } +C$$

haqnatural
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  • This is correct. I do not remember exactly the post to which Simply Beautiful Art replied this morning. –  Feb 01 '17 at 19:22
  • Here is the link. The comment from Simply Beautiful Art is the one to which I was referring. http://math.stackexchange.com/questions/2123271/methods-for-choosing-u-and-dv-when-integrating-by-parts –  Feb 01 '17 at 22:11
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I think you are talking about differentiation under the integral sign. Disregard this post if it is not what you are looking for.

Let us try to evaluate $$\int x e^x \ dx $$ without using integration by parts, the standard method.

Let $F(x,t)=e^{xt}$ and observe $\frac{\partial F(x,t)}{ \partial t} |_{t=1} =x e^x.$

So the integral can be rewritten as

$$\int \frac{\partial F(x,t)}{\partial t} |_{t=1} \ dx= \frac{\partial}{\partial t}|_{t=1} \left (\int F(x,t) \ dx \right)=\frac{\partial}{\partial t}|_{t=1} \left (\int e^{xt}\ dx \right)=\frac{\partial}{\partial t}|_{t=1} \frac{e^{xt} }{t}$$ which is equal to

$$\frac{xe^{xt}-e^{xt}}{t^2}|_{t=1}=xe^x-e^x$$

Vivek Kaushik
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