How can I integrate e^x arctan(x) without Wolfram?

[daw0lf369]

I'm working on a take home exam in my Calculus 2 class. The exam is completely done except for one problem and I desperately need help. I've put so much time into this one problem that I'm ready to just miss it and take the hit.

Homework Statement

Indefinite Integral

∫(e^x)*(arctan(x))dx

Homework Equations

The Attempt at a Solution

I've tried using a couple different substitutions by parts. Using u = arctan(x) results in (e^x)*arctan(x) - ∫(e^x)/(x^2+1) dx I couldn't seem to progress any further on that integral either.

Another attempt used x = tanθ which results in ∫arctan(ln(u))du Once again, I couldn't make any further progress with this integral.



It seems as though the problem was written improperly, but I asked my professor and she said it is correct as written. I would greatly appreciate any and all help you can provide.

 

[1MileCrash]

If that's written correctly, your calc 2 class is hardcore. I doubt that even has an elementary antiderivative, and if it does, it's uglier than gary busey.

 

[daw0lf369]

I'm pretty sure it has to be wrong. The test has 60 problems. The other 59 took me about 10-15 hours total. This one problem has taken at least 8 so far and I'm still no closer to an answer.

 

[DeIdeal]

Well, whenever I have doubts whether there's a mistake on a problem like this, I run it through Wolfram Alpha or an equivalent software and see whether the answer seems like something I'm supposed to be able to do.

Have you tried that? Wolfram seems to give

$\frac{1}{2}ie^{i}Ei(x-i)-\frac{1}{2}ie^{-i}Ei(x+i)+e^{x}\arctan(x)+C$

Does that look familiar?

I have no idea about the syllabus (sorry, is that the right word?) in US (I assume?), or any other countries foreign for me, so I don't know what "Calculus II" would actually cover.

 

[1MileCrash]

i the imaginary unit?

 

[RoshanBBQ]

That solution from Wolfram alpha basically admits it doesn't know the solution. The Ei function is one defined as an integral. But I have encountered integrals for which Wolfram alpha could not solve yet were solvable. It's most likely your teacher selected this challenge problem to stump up students who use automatic solvers.

 

[1MileCrash]

It's most likely your teacher selected this challenge problem to stump up students who use automatic solvers.

..and everyone else.

Enjoy your 59/60.

 

[victor.raum]

Write on the test that you tried like hell, but that you suspect no closed for solution exists. He'll probably enjoy the honesty :-)

 

[1MileCrash]

Maybe it's a trick, and anyone who writes what wolfram said will fail the course.

 

[daw0lf369]

Yet another failed solution. I tried a substitution for e^x and managed to get to ∫1/((ln(x))^2+1)dx

But, once again, that's as far as it'll go.

I'll have to take it into her again and have her show a step or two that she would take. In the past she's been a big stickler for Algebraic manipulation before integration, but I can't get any other forms to fall out of this equation.

Sadly, we haven't done anything similar to adding i into our answers.

 

[daw0lf369]

Just realized this was a double post of sorts

 

[mathstew]

Building of your idea of using integration by parts to obtain:

(e^x)*arctan(x) - ∫(e^x)/(x^2+1)dx

If you consider the function in the integral as a function of a complex variable, you may note that it is holomorphic on open sets U that stay away from i and -i. Using Cauchy's Integral Formula should allow you to construct a primitive on such sets.

But yes, very hard for a Calc II course.

 

[micromass]

But I have encountered integrals for which Wolfram alpha could not solve yet were solvable.

Like what??

That shouldn't be the case actually. If wolfram alpha says that it can't be solved in terms of elementary functions, then it really can't.

 

[DeIdeal]

That solution from Wolfram alpha basically admits it doesn't know the solution. The Ei function is one defined as an integral. But I have encountered integrals for which Wolfram alpha could not solve yet were solvable. It's most likely your teacher selected this challenge problem to stump up students who use automatic solvers.

If wolfram alpha says that it can't be solved in terms of elementary functions, then it really can't.

This, pretty much. I'm familiar with the fact that Wolfram Alpha cannot integrate everything you throw at it, but I've never seen a case where something it managed to integrate could be expressed in a form that's much simpler. Based on this

Sadly, we haven't done anything similar to adding i into our answers.

OP is not familiar with complex-valued integrals, so I don't see how he could have very extensive knowledge when it comes to integrating using special functions. So the result should be greatly simpler than the one above.

(Oh, and just to clarify: If the actual solution is anything like the one Wolfram gave, I wouldn't be able to integrate this myself either, I've only encountered few special functions as of now. It (seeing whether the result is "on your level") was just a trick I think works pretty often, like I said on my previous post.)

 

[Citan Uzuki]

And just to confirm everyone's suspicions, the OP's integral really is nonelementary, and provably so. It can be shown (e.g. on page 971 of http://www.jstor.org/stable/2318066) that if f and g are rational functions, f is nonzero, and g is nonconstant, then the only way that f(x)e^g(x) can have an elementary antiderivative is if it has one of the obvious form a(x)e^g(x), where a is a rational function. So the only way that $e^x(1+x^2)^{-1}$ could have an elementary antiderivative is if there were some rational function a(x) such that:
$$a'(x) + a(x) = \frac{1}{x^2+1}$$

But this is clearly impossible, as any pole of a'+a would have to have order 2 or higher, and 1/(x^2 + 1) has poles of order 1. So e^x/(1+x^2) has no elementary antiderivative. But then e^x arctan(x) can have no elementary antiderivative either, or else one could use it to find an elementary antiderivative of e^x/(1+x^2) using integration by parts. Q.E.D.

 

[jackmell]

$$\int e^{x}\arctan{x}dx=\int\sum_{n=0}^{\infty} \frac{x^n}{n!}\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}dx$$

We can integrate that easily can't we? Is that not Calculus II or no?

 

[RoshanBBQ]

Like what??

That shouldn't be the case actually. If wolfram alpha says that it can't be solved in terms of elementary functions, then it really can't.

I was mistaken with my example. It was actually a definite integral whose limits allowed for a simple solution even though the indefinite integral was very messy in Wolfram Alpha. And on that note, this problem wouldn't happen to have limits, would it?

 

[daw0lf369]

Unfortunately Roshan, it's indefinite.

We've definitely gone over infinite series like packmell suggests, I've seriously considered integrating it that way if I can't find any other option.

Thanks again for all the help, I've submitted another email asking the professor to review the problem. Hopefully she won't blow me off this time.

 

[daw0lf369]

I got the same reply from the Professor, it's correct as written and very challenging but supposedly possible.

With the derivative though, I can't see how you would eliminate things in a way that would leave just e^x*arctan(x).

Although, I was trying a new method and felt like I almost got the answer earlier today. If someone could check to see where I made an error, I would appreciate it.

∫e^x*arctanx dx

let g=e^x ln(g)=x dg=e^xdx

= ∫arctan(lng)dg

by parts u=arctan(lng) dV=dg du=(1/(g(1+ln^2(g))) V=g

= g*arctan(lng)-∫dg/g(1+ln^2(g))

Then another u=ln(g) du=(1/g)dg

= g*arctan(lng) -∫du/(1+u^2)

= g*arctan(lng) - arctan(u) + C

by sub = e^x*arctan(x) - arctan(x) + C

 

[micromass]

I believe your professor is mistaken. This integral does not have an elementary antiderivative.

If you can't find it, be sure to ask your professor his solution. It likely contains an error.

Also, for extra credit, you might want to prove that the integral indeed does not have an elementary antiderivative (not easy!). The main theorem is Liouville's theorem: http://en.wikipedia.org/wiki/Liouville's_theorem_(differential_algebra)

The book "Algorithms for computer algebra" by Geddes is a very good source.

 

[1MileCrash]

= ∫arctan(lng)dg

by parts u=arctan(lng) dV=dg du=(1/(g(1+ln^2(g))) V=g

= g*arctan(lng)-∫dg/g(1+ln^2(g))

No

= g*arctan(lng)-∫gdg/g(1+ln^2(g))

= g*arctan(lng)-∫dg/(1+ln^2(g))

Which wrecks the rest of your work entirely because what you wrote,:

∫dg/g(1+ln^2(g))

Has an elementary antiderivative, while when you correctly multiply Vdu for:

∫dg/(1+ln^2(g))

does not.

 

[daw0lf369]

I figured it was too good to be true. Thanks. The assignment's due friday, I figure once I get it back, I'll ask the Prof for a solution and if she doesn't have one, i'll go over her head for the points. I've spent way to much time and brought it up to her too often to just drop it.

 

[daw0lf369]

Well on Friday we turned in the last half of our take home test. The Professor is very strict and if you don't turn in your test at the beginning of class you're out of luck and get a zero. So after we had turned them in, a classmate asked her to show us the solution to this problem. She flat out said no way, she wanted to be able to use it again in the future. I've never seen a teacher of any sort that was unwilling to actually teach. Because it was obvious to everyone that the problem was unsolvable and she had made a mistake and was unwilling to own up to it.

Anyhow, I wanted to thank everyone on here for the help. I really appreciated the feedback and support.

 

[micromass]

Well on Friday we turned in the last half of our take home test. The Professor is very strict and if you don't turn in your test at the beginning of class you're out of luck and get a zero. So after we had turned them in, a classmate asked her to show us the solution to this problem. She flat out said no way, she wanted to be able to use it again in the future. I've never seen a teacher of any sort that was unwilling to actually teach. Because it was obvious to everyone that the problem was unsolvable and she had made a mistake and was unwilling to own up to it.

Anyhow, I wanted to thank everyone on here for the help. I really appreciated the feedback and support.

That seems wrong. Homework assignments are there to help you learn. If you never see the the correct solutions, then how are you expected to learn from it??
Is there anything you can do against it?? Perhaps talk to another professor or something?? It's up to you of course.

 

[1MileCrash]

Unacceptable.

 

[autodidude]

Though this is way beyond my level, I'm also interested in the solution :p:...unless if it was in fact an error...

 

[RoshanBBQ]

Who knows what the professor is doing. Maybe he just likes to give out problems he knows are impossible just to see who is more driven (i.e. if he receives a lot of feedback from a student, he knows they care more). Or maybe he made an error in his solution.

 

[AliAhmed]

Not that it matters anymore, but the solution requires knowledge in exponential integrals. It does not have an absolute antiderivative. The antiderivative itself must be expressed in terms of exponential integrals. For the user that used Wolfram, the Ei stands for exponential integral.

 

[DeIdeal]

She flat out said no way, she wanted to be able to use it again in the future. I've never seen a teacher of any sort that was unwilling to actually teach. Because it was obvious to everyone that the problem was unsolvable and she had made a mistake and was unwilling to own up to it.

Are you serious?! Damn, I wouldn't want to have a teacher like that. That just sounds downright ridiculous. I'm seconding the suggestion to ask another professor.

For the user that used Wolfram, the Ei stands for exponential integral.

Well, yeah, I know. I just said I don't know how to actually use them (nor do I need to be, at least not yet.)