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I gave an integral to a student. She reported back to me that she could not do it. I've tried a couple of approaches and have failed. I imagine it is fairly easy. It's a double integral. And, no matter, which variable you do first, you end up with an integral like: $$ \int_1^2 \frac{1}{3x}e^{x^5}\,dx. $$ Any help would be appreciated. In case I have messed up the initial integrations, the original integral was: $$ \int_1^2\int_1^3 x^4y^2e^{x^5y^3}\,dy\,dx. $$

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If only the original integrand was $x^4y^2e^{x^5+y^3}$... –  1015 Jan 28 '13 at 19:30

2 Answers 2

up vote 3 down vote accepted

This integral pertains to the exponential integral function, $\text{Ei}(z)$, so you weren't going to get it done by hand.

Mathematica confirms:

Mathematica graphics

Just to make sure, let's check the double integral as well:

Mathematica graphics


PS Note the particular definition Mathematica employs here for $\text{Ei}(z)$:

Mathematica graphics

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Thanks. If you'll believe it, I grabbed the integral from an ordinary undergraduate calculus book. –  Joe Johnson 126 Jan 29 '13 at 1:35
    
Yeah, probably just a typo somewhere... –  JohnD Jan 29 '13 at 1:46

FWIW, here's how $\mathrm{Ei}$ shows up:

$$\int\frac{\exp\,t^5}{t}\mathrm dt=\frac15\int\frac{5t^4 \exp\,t^5}{t^5}\mathrm dt=\frac15\int\frac{\exp\,u}{u}\mathrm du$$

where the simple substitution $u=t^5$ was made. Since $\mathrm{Ei}$ is defined as the integral of $\frac{\exp\,u}{u}$, that's how it turns up.

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