Antiderivative of 1/sqrt(lnx - c)

[rebeka]

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

 

[Icebreaker]

Try substituting u = lnx.

 

[unggio]

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

 

[HallsofIvy]

Hmmm, the substitution u= ln(x)- C, along the lines Icebreaker suggested, changes the integral to \int\frac{e^{u+C}du}{\sqrt{u}} but I don't see any simple way to do that.

In fact, the "obvious" substitution u= \sqrt{ln(x)- C} converts the
integral to 2e^C\int e^{u^2}du which has no elementary antiderivative!

 

[saltydog]

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

\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)}}

I'll try too.

 

[lurflurf]

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

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.
On the verification front
the substitution u=i Sqrt(log(x)-C) gives
\int \frac{1}{\sqrt{\log(x)-C}}dx=-2i e^c\int e^{-u^2}du
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]

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.

Hello Lurflurf. The destination is not the objective but rather the journey.

"Equal rights for special functions."

 

[lurflurf]

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

 

[saltydog]

On the verification front
the substitution u=i Sqrt(log(x)-C) gives
\int \frac{1}{\sqrt{\log(x)-C}}dx=-2i e^c\int e^{-u^2}du
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)

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.

 

[saltydog]

I'm making progress with this. Using Lurflurf's substitution, I obtain:

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

Since:

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

Then:

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

Thus:

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

Now, as long as c\leq ln(x),

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

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?

Erf[ia]=?

 

[saltydog]

I'm making progress with this. Using Lurflurf's substitution, I obtain:

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

Since:

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

Then:

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

Thus:

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

Now, as long as c\leq ln(x),

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

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?

Erf[ia]=?

Alright, I got it by considering the Riemann sum:

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

Thus, applying this to the problem above we have:

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

Now:

<br /> \begin{align*}<br /> Erf \left[i\sqrt{ln(x)-c}\right]&amp;=\frac{2}{\sqrt{\pi}}\int_0^{i\sqrt{ln(x)-c}} e^{-t^2}dt \\<br /> &amp;=\frac{2i}{\sqrt{\pi}}\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt<br /> \end{align}<br />

Thus we finally have:

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

 

[saltydog]

Thus we finally have:

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

Well, wait. That's just substituting a tough integral for a more-tough one. So, let's do this:

Let:

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

Then:

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

See, that's much better.

 

[lurflurf]

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

Erf[ia]=?

Your to quick but I'll say this anyway.
erf(i a)=-i erfi(a)=-i\frac{2}{\sqrt{\pi}}\int_0^a e^{z^2} dz
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)

 

[saltydog]

Your to quick but I'll say this anyway.
erf(i a)=-i erfi(a)=-i\frac{2}{\sqrt{\pi}}\int_0^a e^{z^2} dz
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)

I don't think it's a minus i right? That is:

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

I mean you're better at this than me.

 

[lurflurf]

I don't think it's a minus i right? That is:

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

Drats I was thinking sdrawkcab.
erf(i a)=i erfi(a)
I mixed that up with
erfi(a)=-i erf(i a)
to produce the error.