How can we evaluate the following?
$$\int e^{\frac{y}{x}}\ \mathrm dy$$
An explanation of the answer would be helpful.
The answer I got is $ x e^{y/x}$.
But not sure about the steps used for obtaining the answer...
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With regards to which variable are you integrating? $x$? $y$? Both? If so, in which order? – daOnlyBG Nov 20 '14 at 16:07
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In any event, you forgot the constant! – GFauxPas Nov 20 '14 at 16:08
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integrating with respect to dy – jay Nov 20 '14 at 16:10
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Change variable $y=z x$ and it will becopme easy. – Claude Leibovici Nov 20 '14 at 16:32
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How did you get that answer? If you don't show the steps, we can't tell if they are correct. – robjohn Nov 20 '14 at 17:11
3 Answers
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Since this a task on finding a primitive, you can always check you candidate answer by taking (partial in your case) derivative.
SBF
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$$\begin{align} \int_{- \infty} e^{\frac{y}{x}}dy &= \int_{- \infty}^y e^{\frac{t}{x}}dt \\ &= \int_{- \infty}^y e^{\frac{xu}{x}}d(xu) \\ &= x \int_{- \infty}^{y \over x} e^u d u \\ &= x \vert_{u = {- \infty}}^{y \over x} e^u \\ &= x \ e^{y \over x} \end{align}$$
Thumbnail
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First note that $$\int a^t\ \mathrm dt=\frac{a^t}{\ln a}+C$$ Therefore $$\int e^{\frac{y}{x}}\ \mathrm dy=\int \left(e^{\frac{1}{x}}\right)^y\ \mathrm dy$$ $$=\frac{\left(e^{\frac{1}{x}}\right)^y}{\ln e^{\frac{1}{x}}}+C=xe^{\frac{y}{x}}+C$$
k170
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