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## Main Question or Discussion Point

does anybody know the antiderivative of 1/(sqrt(lnx - (a constant))

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does anybody know the antiderivative of 1/(sqrt(lnx - (a constant))

Icebreaker

Try substituting u = lnx.

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i don't think it is possible to solve that problem. could be wrong though

HallsofIvy

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In fact, the "obvious" substitution [tex]u= \sqrt{ln(x)- C}[/tex] converts the

integral to [tex]2e^C\int e^{u^2}du[/tex] which has no elementary antiderivative!

saltydog

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Your task, should you choose to accept it, is to show how Mathematica comes up with the following:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-\frac{e^{c}\sqrt{\pi}\left(1-Erf\left[\sqrt{c-ln(x)}\right]\right)\sqrt{-c+ln(x)}}{\sqrt{c-ln(x)}}[/tex]

I'll try too.

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-\frac{e^{c}\sqrt{\pi}\left(1-Erf\left[\sqrt{c-ln(x)}\right]\right)\sqrt{-c+ln(x)}}{\sqrt{c-ln(x)}}[/tex]

I'll try too.

Last edited:

lurflurf

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You can verify the result if you like by showing the derivatives of each side are equal. Seeing how Mathematica comes up with it though is much harder and not very helpful. Mathematica does not do integral like a person would. If Integrate had a ShowSteps option what you would see would likely be ridiculuously long and complicated.saltydog said:Your task, should you choose to accept it, is to show how Mathematica comes up with the following:

On the verification front

the substitution u=i Sqrt(log(x)-C) gives

[tex]\int \frac{1}{\sqrt{\log(x)-C}}dx=-2i e^c\int e^{-u^2}du[/tex]

Thus all that is needed to bridge to the mathematica result is to use (under suitable conditions)

sqrt(C-log(x))=i sqrt(log(x)-C)

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saltydog

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Hello Lurflurf. The destination is not the objective but rather the journey.lurflurf said:You can verify the result if you like by showing the derivatives of each side are equal. Seeing how Mathematica comes up with it though is much harder and not very helpful. Mathematica does not do integral like a person would. If Integrate had a ShowSteps option what you would see would likely be ridiculuously long and complicated.

"Equal rights for special functions."

lurflurf

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Of course erf's evil brother erfi could also be used.

saltydog

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Thanks Lurflurf. That's very helpful cus' I wasn't getting anywhere in the Real world. Still not happening for me but I'll continue working on it . . . it's a pleasant ride.lurflurf said:On the verification front

the substitution u=i Sqrt(log(x)-C) gives

[tex]\int \frac{1}{\sqrt{\log(x)-C}}dx=-2i e^c\int e^{-u^2}du[/tex]

Thus all that is needed to bridge to the mathematica result is to use (under suitable conditions)

sqrt(C-log(x))=i sqrt(log(x)-C)

saltydog

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[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-2i e^c \int e^{-u^2}du,\quad u(x)=i\sqrt{ln(x)-c}[/tex]

Since:

[tex]Erf(t)=\frac{2}{\sqrt{\pi}}\int e^{-t^2}dt[/tex]

Then:

[tex]\int e^{-t^2}dt=\frac{\sqrt{\pi}}{2}Erf(t)[/tex]

Thus:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}dx=-i e^c\sqrt{\pi} Erf \left[i\sqrt{ln(x)-c}\right][/tex]

Now, as long as [itex] c\leq ln(x)[/itex],

[tex] Erf\left[i\sqrt{ln(x)-c}\right] [/tex]

will be a pure complex number (prove?) and thus the integral will be a positive real number.

I don't know how to calculate (numerical or otherwise) the value of erf(ai). Can someone please show me?

[tex]Erf[ia]=?[/tex]

saltydog

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Alright, I got it by considering the Riemann sum:saltydog said:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-2i e^c \int e^{-u^2}du,\quad u(x)=i\sqrt{ln(x)-c}[/tex]

Since:

[tex]Erf(t)=\frac{2}{\sqrt{\pi}}\int e^{-t^2}dt[/tex]

Then:

[tex]\int e^{-t^2}dt=\frac{\sqrt{\pi}}{2}Erf(t)[/tex]

Thus:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}dx=-i e^c\sqrt{\pi} Erf \left[i\sqrt{ln(x)-c}\right][/tex]

Now, as long as [itex] c\leq ln(x)[/itex],

[tex] Erf\left[i\sqrt{ln(x)-c}\right] [/tex]

will be a pure complex number (prove?) and thus the integral will be a positive real number.

I don't know how to calculate (numerical or otherwise) the value of erf(ai). Can someone please show me?

[tex]Erf[ia]=?[/tex]

[tex]Erf(ia)=\frac{2}{\sqrt{\pi}}\int_0^{ia} e^{-t^2}dt=\frac{2i}{\sqrt{\pi}}\int_0^a e^{t^2}dt[/tex]

Thus, applying this to the problem above we have:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-ie^c\sqrt{\pi} Erf \left[ i\sqrt{ln(x)-c}\right][/tex]

Now:

[tex]

\begin{align*}

Erf \left[i\sqrt{ln(x)-c}\right]&=\frac{2}{\sqrt{\pi}}\int_0^{i\sqrt{ln(x)-c}} e^{-t^2}dt \\

&=\frac{2i}{\sqrt{\pi}}\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt

\end{align}

[/tex]

Thus we finally have:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

Last edited:

saltydog

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Well, wait. That's just substituting a tough integral for a more-tough one. So, let's do this:saltydog said:Thus we finally have:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]

Let:

[tex]Sa[x,c]=\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt;\quad x>0\quad\text{and}\quad ln(x)\geq c[/tex]

Then:

[tex]\int \frac{dx}{\sqrt{ln(x)-c}}=2e^cSa(x,c)[/tex]

See, that's much better.

lurflurf

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Your to quick but I'll say this anyway.saltydog said:I don't know how to calculate (numerical or otherwise) the value of erf(ai). Can someone please show me?

[tex]Erf[ia]=?[/tex]

[tex]erf(i a)=-i erfi(a)=-i\frac{2}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]

for fun put the integral into mathematica differently and try to get a result in terms of erfi.

like

integral=exp(C)sqrt(pi)efi(sqrt(x))

Also when trying to get the same form as mathematica

u=i sqrt(log(x)-C)=sqrt(-log(x)+C)

Last edited:

saltydog

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I don't think it's a minus i right? That is:lurflurf said:Your to quick but I'll say this anyway.

[tex]erf(i a)=-i erfi(a)=-i\frac{2}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]

for fun put the integral into mathematica differently and try to get a result in terms of erfi.

like

integral=exp(C)sqrt(pi)efi(sqrt(x))

Also when trying to get the same form as mathematica

u=i sqrt(log(x)-C)=sqrt(-log(x)+C)

[tex]erf[i a]=\frac{2i}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]

I mean you're better at this than me.

lurflurf

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Drats I was thinking sdrawkcab.saltydog said:I don't think it's a minus i right? That is:

[tex]erf[i a]=\frac{2i}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]

erf(i a)=i erfi(a)

I mixed that up with

erfi(a)=-i erf(i a)

to produce the error.

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