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First. Please read my code:

In[28]:=
Integrate[Log[1/x - 1]/E^(x*t), {x, 0, 1}, Assumptions -> t > 0]

Out[28]=
-Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t]

In[8]:=
Sum[((-x)*t)^i/i!, {i, 0, Infinity}, Assumptions -> t > 0]

Out[8]=
E^((-t)*x)

In[20]:=
-Sum[(-1)^j*((-2 + 1/x)^j/j), {j, 1, Infinity}]

Out[20]=
Log[(1 - x)/x]

In[19]:=
Sum[(m!/((m - k)!*k!))*(1/x)^(m - k)*(-2)^k, {k, 0, m}, Assumptions -> m > 0]

Out[19]=
(-2 + 1/x)^m

In[25]:=
-Sum[(((-x)*t)^i/i!)*(-1)^j*(1/j)*(j!/((j - k)!*k!))*(1/x)^(j - k)*(-2)^k, 
   {i, 0, Infinity}, {j, 1, Infinity}, {k, 0, j}, Assumptions -> t > 0, 
   Assumptions -> j > 0]

Out[25]=
Log[(1 - x)/x]/E^(t*x)

In[29]:=
Integrate[-Sum[(((-x)*t)^i/i!)*(-1)^j*(1/j)*(j!/((j - k)!*k!))*(1/x)^(j - k)*(-2)^k, 
    {i, 0, Infinity}, {j, 1, Infinity}, {k, 0, j}, Assumptions -> t > 0, 
    Assumptions -> j > 0], {x, 0, 1}, Assumptions -> t > 0]

Out[29]=
-Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t]

In[31]:=
Simplify[-Sum[((j - 1)!*(-2)^k*t^i)/(i!*(j - k)!*k!*(i + k - j + 1)), {i, 0, Infinity}, 
     {j, 1, Infinity}, {k, 0, j}, Assumptions -> t > 0, Assumptions -> j > 0] == 
   -Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t]]

Out[31]=
Sum[((-2)^k*t^i*(-1 + j)!)/((1 + i - j + k)*i!*(j - k)!*k!), {i, 0, Infinity}, 
   {j, 1, Infinity}, {k, 0, j}, Assumptions -> t > 0, Assumptions -> j > 0] == 
  Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t]

As you can see there, the last line didn't give me True output that I expected. Please help me.

Further Information & Question: I want to evaluate integral using series. The integral that i'm talking about: $$\int_0^1 e^{-xt}\ln\left(\frac{1-x}{x}\right)\,\Bbb dx$$ That integral is hard to solve by hand (according to me). So, my idea is taking those two functions in taylor series form, then integrating the term that contains variable $x$. Notice that i have $3$ product of sum which first sum is $e^{-xt}$, second is $\ln(1/x -1)$, and the third is the part of the second since the term of the second series still has binomial expression and i expand it using binomial expansion.

My question is, Since the series form is exactly the same with the function and the output of the original integral gave me a derivative of hypergeometric function, it must be true if my triple series has similar value after integrating both function and series from $0$ to $1$. My final question here is "I want to verify the result and the result has to answer "True". And if there is a mistake on my code, please tell me or correct me. Can you help me? Thanks in advance!

This screenshot, maybe help you to understand my question: enter image description here

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  • 2
    $\begingroup$ To make things simpler, Integrate[Exp[-x*t]*Log[(1-x)/x], {x,0,1}] == -Sum[(-t)^k*HarmonicNumber[k]/(k+1)!, {k, 1, ∞}] gives True, for a single-parameter sum formulation of the hypergeometric derivative. $\endgroup$ – Roman Jan 21 at 11:18
  • $\begingroup$ @Roman Nice! Never heard of harmonic number. Btw, is that $\zeta(1)$? And do you know related link about it so maybe i can use that harmonic number instead? $\endgroup$ – user516076 Jan 21 at 12:14
  • $\begingroup$ Yes, harmonic numbers are related to the Riemann zeta function $\zeta(1)$ when not summing all the way to infinity. Not sure how that is useful here though. $\endgroup$ – Roman Jan 21 at 17:36
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    $\begingroup$ @Roman your sum is given by (ExpIntegralEi[-t]+Exp[-t]ExpIntegralEi[t]-(1+Exp[-t])(Log[t]+EulerGamma))/t $\endgroup$ – Andreas Jan 21 at 19:40
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    $\begingroup$ Differentiating your original integral under the integral sign yields -Integrate[x Log[(1 - x)/x] E^(-t x), {x, 0, 1}], which readily evaluates to a combination of hyperbolic sine and cosine integrals. Integrating the result of that with respect to t will then be equivalent to your original integral. $\endgroup$ – J. M.'s ennui Jan 30 at 10:02
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This is not the answer to your question, but to get rid of the derivative from the hypergeometric function.

f = Integrate[
LaplaceTransform[Log[a*(1 - x)/x]/E^(x*t), a, s], {x, 0, 1}, 
Assumptions -> {s > 0, t > 0}] // Simplify // Expand
ff = Limit[InverseLaplaceTransform[f, s, a], a -> 1, Assumptions -> {t > 0}](*Solution is True when t > 0 *)

(*EulerGamma/t + (E^-t EulerGamma)/t + ExpIntegralE[1, t]/t - (
E^-t ExpIntegralEi[t])/t + Log[t]/t + (E^-t Log[t])/t*)

Plot[{-Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t], ff}, {t, 0, 
10}, PlotStyle -> {Red, {Dashed, Black}}, PlotRange -> All, 
PlotLabels -> {"derivative of hypergeometric function", 
"my function"}]

enter image description here

FullSimplify[
EulerGamma/t + (E^-t EulerGamma)/t + ExpIntegralE[1, t]/t - (
E^-t ExpIntegralEi[t])/t + Log[t]/t + (E^-t Log[t])/
t == -Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t], t > 0](*Can't !*)

Plot[{-Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t] - ff}, {t, 0, 
1}](*Errors.Looks True :)*)
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  • $\begingroup$ Hi, thanks for your answer. Btw did u use laplace transform to perform the integral and compare it with hypergeometric function? $\endgroup$ – user516076 Jan 21 at 12:25
  • $\begingroup$ Both answers with ExpIntegralEi[t] and with Derivative[1, 0, 0][Hypergeometric1F1][1, 2, -t are not satisfactory. The fist one reduces the integral under consideration to another integrals and the second one is in terms of a series (or even its analytic continuation). $\endgroup$ – user64494 Jan 21 at 12:27
  • $\begingroup$ @user64494 bcz i don't think it has elementary function representation? Or is it? $\endgroup$ – user516076 Jan 21 at 12:35
  • $\begingroup$ @user64494 , what is satisfactory answers? $\endgroup$ – Mariusz Iwaniuk Jan 21 at 21:57

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