The question asks for the integral of $e^{mx^{3}}.$ How do I calculate the antiderivative of a function that has $e^{x^{3}}$ lets say. I want to make sure I know how to take the antiderivative of exponential functions.
Integral of $e^{mx^{3}}$
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$\begingroup$
calculus
integration
exponential-function
indefinite-integrals
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4Is it $e^m x^3$, $\left(e^{mx}\right)^3$ or $e^{mx^3}$? Exponentiation is not associative, hence e^x^3 is an ambiguous notation. – 2017-02-01
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0Its the third option you put – 2017-02-01
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0Straight from *Mathematica*: $-\frac{x \Gamma \left(\frac{1}{3},-m x^3\right)}{3 \sqrt[3]{-m x^3}}$ – 2017-02-01
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0@DavidG.Stork Lol, you took the easy route to that answer. – 2017-02-01
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0Thanks for all the replies! I dont think my teacher meant to put this in the review. I haven't learned gamma yet idk what that is lol – 2017-02-01
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0Repeated question: http://math.stackexchange.com/questions/270721/how-to-evaluate-the-integral-int-ex3dx – 2017-02-01
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0@Chris You can do it with the exponential integral if that's what you meant. – 2017-02-01
1 Answers
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Let $mt^3=-u,$
$$\int_0^xe^{mt^3}\ dt=-\int_0^{mx^3}\frac1{3\sqrt[3]{mu}}e^{-u}\ du=-\frac1{3\sqrt[3]{m}}\gamma\left(\frac23,mx^3\right)$$
where $\gamma(a,b)$ is the lower incomplete gamma function.