Um is this right:

The integral of e^(1/4 x^2 s) dx

= e^(s x^2 /4) x?

Thank you

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- May 26th 2009, 04:29 AMgconfusedintegrating exponential
Um is this right:

The integral of e^(1/4 x^2 s) dx

= e^(s x^2 /4) x?

Thank you - May 26th 2009, 04:42 AMcraig
- May 26th 2009, 04:44 AMgconfused
Thanks!! Coz i tried writing it up in mathematica to check my answer, and it gave me that and that just looked dodgy. yeahh must have typed it wrong... thanks again :)

- May 26th 2009, 04:49 AMcraig
Just a little tip on integrating exponentials.

If the question is $\displaystyle \int e^{f(x)} dx$

Then the answer is $\displaystyle \frac{1}{f'(x)} e^{f(x)} + C$

For example, $\displaystyle \int e^{9x} dx$ would be $\displaystyle \frac{1}{9} e^{9x} + C$

You can of course check this by differentiating your answer and see if you get your original equation.

NOTE: This only works if the f(x) is a linear equation. - May 27th 2009, 04:13 AMmr fantastic
Sorry but if that's the question then this answer is totally wrong. Differentiate it and see for yourself.

$\displaystyle \int e^{\frac{sx^2}{4}} dx$ has no elementary primative.

Sorry but in general this is totally wrong. It is only correct when f(x) is a linear function. - May 27th 2009, 04:18 AMgconfused
- May 27th 2009, 04:20 AMcraig
- May 27th 2009, 04:55 AMmr fantastic
I mean that the answer to $\displaystyle \int e^{\frac{sx^2}{4}} \, dx$ cannot be expressed as a finite number of elementary functions. If you post the details of your calculation I will point out your mistakes. Did you try differentiating your answer? Do you get $\displaystyle e^{sx^2/4}$?

If this integral arises as a result of a question you're trying to answer, I suggest you post the entire question exactly as it's written. - May 27th 2009, 05:04 AMHallsofIvy
It is not correct because the derivative of $\displaystyle \frac{2}{sx}e^{sx^2/4}$ has to be done by the product rule: (fg)'= f'g+ fg'. The derivative of $\displaystyle e^{sx^2/4}$ is $\displaystyle \frac{2sx}{4}e^{sx^2/4}$ and that, multiplied by $\displaystyle \frac{2}{sx}$

**will**give $\displaystyle e^{sx^2/4}$ but then the derivative of $\displaystyle \frac{2}{sx}$ is $\displaystyle -\frac{2}{sx^2}$ so the derivative of the entire $\displaystyle \frac{2}{sx}e^{sx^2/4}$ is $\displaystyle e^{sx^2/4}- \frac{2}{sx^2}e^{sx^2/4}$, not just $\displaystyle e^{sx^2/4}$.

Craig's advice for integrating $\displaystyle e^{f(x)}$ in general is not correct because what he is**trying**to do is use the substitution u= f(x). Then du= f'(x)dx so dx= du/f'(x). Putting those into the integral we have $\displaystyle \int e^u\left(du/f'(x)\right)$ and, apparently, he is taking that f'(x) outside the integral. But if f is not linear, then f' is a function of x and**cannot**be taken outside the integral- it must be integrated also.**If**f were linear, that is, if f(x)= ax+ b then f'(x)= a, a constant and we can do that: Letting u= ax+ b, du= adx so dx= dx/a and $\displaystyle \int e^{ax+ b}dx= \int e^u (du/a)= \frac{1}{a}\int e^u du$$\displaystyle = \frac{1}{a}e^u+ C= \frac{1}{a}e^{ax+b}+ C$. But that is**only**possible with a being a constant.

You should understand that**most**(in a very specific sense "almost all") integrable functions do NOT have "elementary functions" as integrals. (Elementary functions" are basically all the functions that you see in Calculus or lower level courses.)

About the best you can do with $\displaystyle \int e^{sx^2/4}dx$ is make the**linear**substitution $\displaystyle u= \sqrt{s}x/2$. Then $\displaystyle du= \sqrt{s}/2 dx$ and $\displaystyle dx= \frac{2}{\sqrt{s}} du$ so $\displaystyle \int e^{sx^2/4} dx = \frac{2}{\sqrt{s}} \int e^{u^2} du = \frac{2}{\sqrt{s}} \text{erf}(x)$ where "erf(x)" is**defined**as $\displaystyle \int e^{u^2}du$. It is a non-elementary function, called the "error function" and is used extensively in statistics. In fact, I wouldn't be surprised if that problem came from a statistics course.