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Anti-derivative of $frac{exp(x)-1}{x}$


Proving that $exp(z_1+z_2) = exp(z_1)exp(z_2)$ with power seriesIntegrate $int_{-infty}^{infty}expleft(-frac{pi^2t(2x+1)^2}{2c^2}right)cosleft(frac{(2x+1)pi y}{c}right)exp(-2pi i kx)dx$Integral of exp(a/x)How to find the area for the curve $y=sin^3(2x)cos^3(2x)$?Why does $zmapsto exp(-z^2)$ have an antiderivative on $mathbb C$?Trouble with indefinite integral $ int sqrt{csc x-sin x} dx $computation of $int_{0}^{infty}frac{1}{theta^{2n+1}}expleft(- frac1{theta^2}sum_{i=1}^n x_i^2right)dtheta$How to find $b_n$ for the limit comparison test in $sum_{n=1}^{infty}frac{(ln(n))^2}{sqrt{n}(10n-9sqrt{n})}$.I need help calculating the anti derivative of $f(x)=2* frac{c^{2x+1}}{2x+1}$Computing $I'(x)$ of $I(x)=int_{-exp(x)}^{x^2}cos(xt^2) dt$













0












$begingroup$


I am looking for the antiderivative of $$frac{exp(x)-1}{x}$$ I showed that it is equivalent to calculate $$sum_{n=1}^{infty}frac1n frac{x^n}{n!}$$ but I can't find both of the solutions. If someone could help me I would very appreciate it. Thanks!






calculus integration power-series indefinite-integrals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
    $endgroup$
    – egreg
    Mar 10 at 15:01






  • 4




    $begingroup$
    You can write the solution in terms of this non-elementary function.
    $endgroup$
    – J.G.
    Mar 10 at 15:03
















0












$begingroup$


I am looking for the antiderivative of $$frac{exp(x)-1}{x}$$ I showed that it is equivalent to calculate $$sum_{n=1}^{infty}frac1n frac{x^n}{n!}$$ but I can't find both of the solutions. If someone could help me I would very appreciate it. Thanks!






calculus integration power-series indefinite-integrals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
    $endgroup$
    – egreg
    Mar 10 at 15:01






  • 4




    $begingroup$
    You can write the solution in terms of this non-elementary function.
    $endgroup$
    – J.G.
    Mar 10 at 15:03














0












0








0





$begingroup$


I am looking for the antiderivative of $$frac{exp(x)-1}{x}$$ I showed that it is equivalent to calculate $$sum_{n=1}^{infty}frac1n frac{x^n}{n!}$$ but I can't find both of the solutions. If someone could help me I would very appreciate it. Thanks!






calculus integration power-series indefinite-integrals







share|cite|improve this question











$endgroup$




I am looking for the antiderivative of $$frac{exp(x)-1}{x}$$ I showed that it is equivalent to calculate $$sum_{n=1}^{infty}frac1n frac{x^n}{n!}$$ but I can't find both of the solutions. If someone could help me I would very appreciate it. Thanks!






calculus integration power-series indefinite-integrals




calculus integration power-series indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 10 at 15:43







Odys

















asked Mar 10 at 14:56









OdysOdys

11




11








  • 3




    $begingroup$
    For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
    $endgroup$
    – egreg
    Mar 10 at 15:01






  • 4




    $begingroup$
    You can write the solution in terms of this non-elementary function.
    $endgroup$
    – J.G.
    Mar 10 at 15:03














  • 3




    $begingroup$
    For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
    $endgroup$
    – egreg
    Mar 10 at 15:01






  • 4




    $begingroup$
    You can write the solution in terms of this non-elementary function.
    $endgroup$
    – J.G.
    Mar 10 at 15:03








3




3




$begingroup$
For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
$endgroup$
– egreg
Mar 10 at 15:01




$begingroup$
For $xne0$, this is equivalent to compute the antiderivative of $e^x/x$, which cannot be expressed in terms of elementary functions.
$endgroup$
– egreg
Mar 10 at 15:01




4




4




$begingroup$
You can write the solution in terms of this non-elementary function.
$endgroup$
– J.G.
Mar 10 at 15:03




$begingroup$
You can write the solution in terms of this non-elementary function.
$endgroup$
– J.G.
Mar 10 at 15:03










1 Answer
1






active

oldest

votes


















2












$begingroup$

As long as you cannot use the exponential integral function, leave it as you wrote
$$intfrac{e^x-1}{x},dx=sum_{n=1}^{infty}frac1n frac{x^n}{n!}>sum_{n=1}^{infty}frac1{n+1} frac{x^n}{n!}=frac 1xsum_{n=1}^{infty} frac{x^{n+1}}{(n+1)!}=frac{e^x-x-1}x$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok, thank you for your answer !
    $endgroup$
    – Odys
    Mar 11 at 16:33












  • $begingroup$
    @Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Mar 11 at 17:22











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

As long as you cannot use the exponential integral function, leave it as you wrote
$$intfrac{e^x-1}{x},dx=sum_{n=1}^{infty}frac1n frac{x^n}{n!}>sum_{n=1}^{infty}frac1{n+1} frac{x^n}{n!}=frac 1xsum_{n=1}^{infty} frac{x^{n+1}}{(n+1)!}=frac{e^x-x-1}x$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok, thank you for your answer !
    $endgroup$
    – Odys
    Mar 11 at 16:33












  • $begingroup$
    @Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Mar 11 at 17:22
















2












$begingroup$

As long as you cannot use the exponential integral function, leave it as you wrote
$$intfrac{e^x-1}{x},dx=sum_{n=1}^{infty}frac1n frac{x^n}{n!}>sum_{n=1}^{infty}frac1{n+1} frac{x^n}{n!}=frac 1xsum_{n=1}^{infty} frac{x^{n+1}}{(n+1)!}=frac{e^x-x-1}x$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok, thank you for your answer !
    $endgroup$
    – Odys
    Mar 11 at 16:33












  • $begingroup$
    @Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Mar 11 at 17:22














2












2








2





$begingroup$

As long as you cannot use the exponential integral function, leave it as you wrote
$$intfrac{e^x-1}{x},dx=sum_{n=1}^{infty}frac1n frac{x^n}{n!}>sum_{n=1}^{infty}frac1{n+1} frac{x^n}{n!}=frac 1xsum_{n=1}^{infty} frac{x^{n+1}}{(n+1)!}=frac{e^x-x-1}x$$






share|cite|improve this answer









$endgroup$



As long as you cannot use the exponential integral function, leave it as you wrote
$$intfrac{e^x-1}{x},dx=sum_{n=1}^{infty}frac1n frac{x^n}{n!}>sum_{n=1}^{infty}frac1{n+1} frac{x^n}{n!}=frac 1xsum_{n=1}^{infty} frac{x^{n+1}}{(n+1)!}=frac{e^x-x-1}x$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 10 at 15:57









Claude LeiboviciClaude Leibovici

124k1157135




124k1157135












  • $begingroup$
    ok, thank you for your answer !
    $endgroup$
    – Odys
    Mar 11 at 16:33












  • $begingroup$
    @Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Mar 11 at 17:22


















  • $begingroup$
    ok, thank you for your answer !
    $endgroup$
    – Odys
    Mar 11 at 16:33












  • $begingroup$
    @Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
    $endgroup$
    – Claude Leibovici
    Mar 11 at 17:22
















$begingroup$
ok, thank you for your answer !
$endgroup$
– Odys
Mar 11 at 16:33






$begingroup$
ok, thank you for your answer !
$endgroup$
– Odys
Mar 11 at 16:33














$begingroup$
@Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
$endgroup$
– Claude Leibovici
Mar 11 at 17:22




$begingroup$
@Odys. You are welcome ! You did a good job. Sooner or later, you will learn a lot about special functions and ... you will enjoy them. Cheers :-)
$endgroup$
– Claude Leibovici
Mar 11 at 17:22


















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