Indefinite integral of $$\int \tan (x) e^ {(\tan (x))} \, dx$$ I have tried using integration by parts but I couldn't integrate it.
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Why do you think that this integral could be evaluated? – user600016 Oct 05 '19 at 11:38
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1The result should be $$\frac{1}{2} e^{-i} \left(e^{2 i} \text{Ei}(\tan (x)-i)+\text{Ei}(\tan (x)+i)\right)$$ – Dr. Sonnhard Graubner Oct 05 '19 at 11:40
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This question was given in my book. – Ayus Das Oct 05 '19 at 11:41
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1@Dr.SonnhardGraubner how are we getting that result? – Ayus Das Oct 05 '19 at 11:42
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Let $\tan(x)=y$ making $$I=\int \tan (x) e^ {\tan (x)} \, dx=\int \frac y{1+y^2} e^y \,dy=\int \frac y{(y+i)(y-i)} e^y \,dy$$
Now, partial fraction decomposition and some obvious changes of variable will take you to the definition of the exponential integral function.
$$I=\frac12 \left(\int\frac{e^y}{ y+i}dy+\int\frac{e^y}{ y-i}dy\right)$$
$$J=\int\frac{e^y}{ y+a}dy=e^{-a}\int\frac{e^{y+a}}{ y+a}dy=e^{-a}\int\frac{e^{t}}{ t}dt=e^{-a}\,\text{Ei}(t) $$ So, $$I=\frac12 \left(e^i\, \text{Ei}(\tan (x)-i)+e^{-i}\,\text{Ei}(\tan (x)+i) \right)$$
Claude Leibovici
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