0
$\begingroup$

Recently I asked a question where as answer came out $\ln{(a)}=2*\text{arctanh } u$ which is to be calculated using $\sum_{k=0}^{\infty}(\frac{c^{2k+1}}{2k+1})$ with $u=\frac{a-1}{a+1} < 1$ if that helps.

Now since I need to approach and not necessarily calculate $\ln{a}$, I figured that if I used it like the Riemann-sums, the antiderivative of the function can also be used, since the sum is leading to an area of $1*f(x)$ and the antiderivative will approximately reach that value (not quite exact, but like I said, that was not necessary).

Anyways, my question was, what is the antiderivative of $f(x)=2* \frac{c^{2x+1}}{2x+1}$? I don't really know if it's very hard, I'm just very bad at calculating antiderivatives when they're not as simple :s

$\endgroup$
  • 1
    $\begingroup$ Subbing $t:=2x+1$ and $\mathrm dt=2~\mathrm dx$, we have $\int\frac{c^t}t~\mathrm dt$. Further subbing $p:=\log t$ and $\mathrm dp=\frac{\mathrm dt}t$ gives $\int c^{e^p}~\mathrm dp=\operatorname{Ei}(e^x\log c)+C$ $\endgroup$ – Prasun Biswas Dec 26 '17 at 9:51
  • 1
    $\begingroup$ Edit: Should be $\operatorname{Ei}(e^p\log c)$ since $e^{\log(c^{e^p})}=e^{e^p\log c}$. Simplifying should give the answer to be $\operatorname{Ei}((2x+1)\log c)+C$ with C, an arbitrary constant. $\endgroup$ – Prasun Biswas Dec 26 '17 at 10:00
  • $\begingroup$ Using 2 artanh u = log ((1 + u)/(1 - u)) might be easier. $\endgroup$ – random Dec 26 '17 at 10:28
5
$\begingroup$

I’m afraid the solution isn’t elementary. But,

$$I = \int 2 \frac{c^{2x+1}}{2x+1}\, dx = \int \frac{c^u}{u}\, du$$ substituting $u = 2x+1$. Now, this is an exponential integral given by: $$I = \operatorname{Ei}\bigl((2x+1)\ln c\bigr) + C$$

$\endgroup$
  • $\begingroup$ I think it should be $\operatorname{Ei}(e^x\ln c)+C$ $\endgroup$ – Prasun Biswas Dec 26 '17 at 9:53
  • 1
    $\begingroup$ WA says it’s fine, @PrasunBiswas. $\endgroup$ – Rohan Dec 26 '17 at 9:54
  • 1
    $\begingroup$ Oh right, I made a mistake computing $\operatorname{Ei}(e^p\ln c)=\operatorname{Ei}(e^{\ln(2x+1)}\ln c)=\operatorname{Ei}((2x+1)\ln c)$ $\endgroup$ – Prasun Biswas Dec 26 '17 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.