Now
So we now need to evaluate
The first term is obviously
Now let us examine the other terms
This can easily be gotten by letting
So we have
And
Similarly with a substitution of we get
So now putting this all back together we get
So seeing that
We need to find
and
So we see that
and
We put this together to get
We can conclude that
Hi! Here are 2 improper integrals for you!
1:
Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo
2:
Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo
i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit
Abs[x] - is the absolute value of x
I would be thankful, if s.o. could type both integrals with TeX, so that I can check if everything's like it should be
I will attempt to LaTex your integrals. I hope I am interpreting correctly.Hi! Here are 2 improper integrals for you!
Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo
I will use k instead of i.Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo, i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit
The function under the integral sign is even. Then we can write :
Since x is positive, we can write :
Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :
Substitute
x=sqrt(u) --> ln(x)=ln(u)/2
The integral is now (after some simplifcations) :
And now, I think an integration by parts can be done, but there's something worrying me
I haven't checked it, but assuming your work up to the last step is corect you cannot do integration by parts because natural log is not differentiably redundant. But what you can do is convert e^{-ax} to power series. Assume uniform convergence and switch integral and summation, do integratioy by parts on [math ]\int_0^1x^n\ln(x)~dx[/tex] and then go from there.
You got it?
Lets look at the first integral. If we let
then we see that So we get
Now assuming uniform convergence
Now for the second integral once again letting
We get
So putting it all together we get
I don't care enough to simplify, its too late. I am not sure that is right, its hard to switch back and forth from that Moo website to MHF and not make clarical errors. And this is all assuming Moo's integral was teh correct one.
By the way in the step before the solution I didn't type a negative in front of the serise and there was no way i was going back to change it, its fixed in the final solution. And gamma should be negative
Not seen in this thread, maybe in another one...?
*code*
Substitute ~ x=\tan \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(1+\tan \varphi\right) ~d \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(\cos \varphi+\sin \varphi\right)- \ln(\cos \varphi) ~d \varphi \\ \\
Let~J=\int_0^{\frac \pi 4} \ln(\cos \varphi+\sin \varphi)~d\varphi \\
We~can~observe~that~ \sin \left(x+\frac \pi 4 \right)=\frac{1}{\sqrt{2}}(\cos(x)+\sin(x)) \\
Thus~\cos \varphi+\sin \varphi=\sqrt{2} \sin \left(\varphi+\frac \pi 4\right) \\ \\
J=\int_0^{\frac \pi 4} \ln(\sqrt{2})+\ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
J=\ln(2) \frac{ \pi }{ 8 } + \int_0^{\frac \pi 4} \ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
Let~\phi=\frac \pi 4 - \varphi \Rightarrow \varphi+\frac \pi 4=\frac \pi 2-\phi \\
J=\ln(2) \frac \pi 8 -\int_{\frac \pi 4}^0 \ln \left(\sin \left(\frac \pi 2-\phi\right)\right)~d \phi \\ \\
J=\ln(2) \frac \pi 8+\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi \\ \\ \\
Therefore \ I=J-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d \varphi \\
I=\ln(2) \frac \pi 8+\underbrace{\int_0^{\frac \pi 4} \ln(\cos \phi) \ d\phi-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d\varphi}_{=0} \\ \\
I=\ln(2){\frac \pi 8}
I must prove this so I can use it later
This is very simple first let us consider a different integral
Now let
then
Now we see that the limits are changed
So we get
Now let
Limits stay the same. So we get
Now notice that
So now dividing both sides appropriately gives us the final result