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I know for a fact that nothing in my expression is wrong, then why cant Wolfram|Alph graph:

sum_(n=0)^∞ (-(ExpIntegralE[-n, (-1 + n) Log[x]] Log[x]^(1 + n))) ? 

Note: The above expression is in Wolfram L anguage code.

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closed as off-topic by corey979, m_goldberg, Henrik Schumacher, MarcoB, Bob Hanlon Feb 11 at 0:57

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    $\begingroup$ sum_(n) etc is not valid syntax in Mathematica -- it looks more like latex. $\endgroup$ – bill s Feb 10 at 18:57
  • $\begingroup$ @bills This is wolfram language code $\endgroup$ – Rithik Kapoor Feb 10 at 18:59
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    $\begingroup$ @RithikKapoor I was solely referring to your statement "This is wolfram language code". Of course, WolframAlpha is rather tolerant towards input. $\endgroup$ – Henrik Schumacher Feb 10 at 19:39
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    $\begingroup$ I'm voting to close this question as off-topic because the tag description of "wolfram-alpha-queries" says "Use this for questions about calling Wolfram Alpha from Mathematica. Questions about Wolfram Alpha itself are off-topic". $\endgroup$ – corey979 Feb 10 at 20:01
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    $\begingroup$ I agree with @HenrikSchumacher, Wolfram alpha does not use the Wolfram language for input and it has it own AI based rules. Here is an example where it returns the state of texas as an answer for an integrate command, because the integrate command was missing the integration variable. This will not happen if using the Wolfram language as input using Mathematica as it will return a syntax error. So answering Wolfram alpha syntax questions means it is out of scope for Wolfram language used by Wolfram Mathematica. $\endgroup$ – Nasser Feb 10 at 22:25
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Assuming that you mean

Sum[-ExpIntegralE[-n, (n-1) Log[x]] Log[x]^(1 + n), {n, 0, ∞}]

the term $n=1$ is ComplexInfinity, and so the entire formula is infinite.

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  • $\begingroup$ No it is not as the following expression is equal to $ \int \sqrt[x]{x} dx $. I think only someone with pro computation time can plot this. $\endgroup$ – Rithik Kapoor Feb 10 at 19:14
  • $\begingroup$ $\lim_{z\to0} E_{-1}(z)=+\infty$ is what's relevant for $n=1$. You can see this from $E_{-1}(z)=e^{-z}(z+1)/z^2$, which you get in Mathematica with ExpIntegralE[-1, z] // FunctionExpand // FullSimplify. $\endgroup$ – Roman Feb 10 at 19:39
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If you can't get WolframAlpha, or anything else for that matter, to plot an infinite sum over the reals then see if it can plot a finite sum over some points that is close enough to what you want to see.

This

plot Table[sum_(n=2)^10 (-(ExpIntegralE[-n, (-1+n) Log[x]] Log[x]^(1+n))),{x,2.,100.,.25}]

plots just fine under WolframAlpha and the upper bound on the summation seems to be enough to show the behavior of the expression. Likewise the range of x.

Notice I skipped 0 and 1 to avoid the infinities.

People need to make it really really REALLY clear that they are asking a question ONLY within the domain of WolframAlpha or they will almost always get people who will answer something about Mathematica instead. And that will often happen even when people do make it REALLY clear. That happens even with WolframAlpha questions posted to https://community.wolfram.com/

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