Is this integral solvable? $$\int \frac{e^x}{x^2-a^2}dx,\quad a>0.$$
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2The anti-derivative is not an elementary function. – mrf Apr 15 '14 at 19:13
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2It is not an elementary integral. You can have a closed form solution in terms of the exponential integral. – Mhenni Benghorbal Apr 15 '14 at 19:13
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2How do you know? Proof? – user85798 Apr 15 '14 at 19:20
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Is this true? $\int\frac{e^x}{x^2-a^2}dx=\frac{e^{a+i(x-a)}-e^{-a+i(x+a)}}{2a}$. How? – David Apr 15 '14 at 19:21
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It involves this: http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/ so, in other words, not solvable. – Shahar Apr 15 '14 at 19:27
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1@David: The answer is in terms of the "exponential integral function" not "the exponential function". See here. – Mhenni Benghorbal Apr 15 '14 at 19:37
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Yeah! You are right. – David Apr 16 '14 at 04:55
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$$x^2-a^2=(x-a)(x+a)\quad\Longrightarrow\quad\dfrac1{x^2-a^2}=\dfrac1{2a}\left(\dfrac1{x-a}-\dfrac1{x+a}\right)$$
\begin{alignat*}{9} \int\frac{e^x}{x^2-a^2}dx\ &=\dfrac1{2a}\bigg(&&\int\dfrac{e^x}{x-a}dx&-&&&\int\dfrac{e^x}{x+a}dx&\bigg)\\ &=\dfrac1{2a}\bigg(&e^a&\int\dfrac{e^{x-a}}{x-a}dx&-&&\ e^{-a}&\int\dfrac{e^{x+a}}{x+a}dx&\bigg)\\ &=\dfrac1{2a}\bigg[&e^a&\int\dfrac{e^{x-a}}{x-a}d(x-a)\ &-&&\ e^{-a}&\int\dfrac{e^{x+a}}{x+a}d(x+a)&\bigg]\\ &=\frac1{2a}\Big[&e^a\,&\text{Ei}(x-a)&-&&\ e^{-a}\,&\text{Ei}(x+a)&\Big] \end{alignat*}
where Ei$(x)$ is the exponential integral, which is not expressible in terms of elementary functions. See Liouville's theorem and the Risch algorithm for more details.
Lucian
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