# Any tips to solve this integral : $I_1 = \int \ln(x^2)e^{\sin(x)}\sin(x^{\cos(x)}) dx$

Background: I was making new expressions to see whether I could efficiently find their derivatives... After having done that, I've started trying to integrate most of them; obviously most of them don't seem to have closed forms..

I need some help in solving the below integral... Substitutions (Trig and Weiertrass) seem to be ineffective; maybe I'm not finding out the right substitutions ...

$$I_1 = \int \ln(x^2)e^{\sin(x)}\sin(x^{\cos(x)}) dx$$

(Note: Solution can have special functions eg: Bessel functions, Hypergeometric, etc; not necessary to be in closed form.)

• On what basis do you believe that there will be a closed form for either of these two expressions? – Semiclassical Feb 24 '15 at 3:47
• No where have I stated that I believe these have closed forms (i.e. I haven't said anything about the solutions being necessarily closed form; solutions may contain functions like the error function, gamma function, for example, which makes a solution no longer in a close form, rather an analytical expression(as opposed to a closed form expression; en.wikipedia.org/wiki/Closed-form_expression) Essentially, I don't know whether these have a closed form solution or not. I just wished to know if there are possible solutions to these out of curiosity. Do you have any idea? @Semiclassical – Kugelblitz Feb 24 '15 at 3:54
• Why the sudden downvote? @downvoter Anything I can do to make the question better? – Kugelblitz Feb 24 '15 at 4:21
• Well the first thing I would do in regards to the first integral is break the integral up to avoid the nasty absolute value – ASKASK Feb 24 '15 at 6:21
• @ASKASK I tried to, but wasn't successful; will try again. – Kugelblitz Feb 24 '15 at 6:27

This is not an answer but my point of view.

First of all contour integration DOES not work here. Reason is that we don't have an improper integral of the form $\displaystyle \int_{0}^{\infty}\ln x^2 e^{\sin x}\sin \left ( x^{\cos x} \right )\,{\rm d}x$. These kind of integrals can be killed with contour integration ,that is integrals of the form:

$$\int_{-\infty}^{\infty}f(x)\, dx, \;\;\;\;\int_{0}^{\infty}f(x)\,dx$$

because we have to let the radius or the side of the rectangle tend to $+\infty$.

So, this method is not going to help us at all!

My strong belief is that these integrals DO not have a closed form. I have seen integrals of the form $\displaystyle \int e^{\cos x}\, dx, \;\; \int e^{\sin \left ( \cos x \right )}\, dx$ being evaluated in terms of Bessel functions (I don't know these functions at all ) so clearly I know that some weird integrals like these before can be tackled using these kind of function.

But here this $\ln x^2$ cause a lot of trouble.The only thing one can do is to drop the square . It should be pointed out that $I_1$'s interval of integration is $\mathbb{R}^*$ and the function is continuous there but indefinite integration is only valid in interval not union of intervals. So, someone is going to integrate over either $(0, +\infty)$ or $(-\infty, 0)$.

$\bullet$ $x \in (0, +\infty)$ $\displaystyle \Rightarrow \int \ln x^2 e^{\sin x}\sin \left ( x^{\cos x} \right )\,dx =2\int \ln x \,e^{\sin x}\sin x\left ( x^{\cos x} \right )\,dx$

Now, note that the integrand is continuous in $(0, +\infty)$ (quite clearly). That means that there exists a primitive (that is a differentiable function $F$ such that $F'=f$, whereas $f$ is the integrand). We can express the primitive as a definite integral , say $\displaystyle F(x)=2\int_1^x \ln t \,e^{\sin t}\sin t\left ( t^{\cos t} \right )\,dt$ but we cannot evaluate it because of that $\ln$ term. I tried out parts and it gets nastier. Also, there exists no obvious substitution that could get things running.

The same comment I have for the case that $x \in (-\infty, 0)$. You can define $F$ as: $$F(x)=-2\int_{-1}^x \ln (-t) \,e^{-\sin t}\sin t\left ( t^{\cos t} \right )\,dt$$

and as one can see $F$ is indeed differentiable. Of course you can play with the lower limit and produce as many primitives as you wish.

As for $I_2$ this is going to be more problematic, first of all because of the $\ln x^2$ term and second of all because of the denominator. Again you're integrating over an interval so that you are able to get rid of the absolute value, but even with that things are quite nasty not because of the trigonometric functions involved but of the $\ln x^2$ as I said before.

Maybe if we get rid of the $\ln x^2$ we can get nice closed form for the second integral, involving or not special functions but as it is my intuition says it can just not be done!

Besides do not forget that every integral you meet does not always have closed form. Differentiation of elementary function is always an elementary function, but integration is not. The examples are many and quite known , therefore I do not write something down.

• I'm well aware of the fact that not all integrals have closed forms or solutions with elementary functions. Also, thank you for explaining why contour integration won't work, and why the integral most probably doesn't have a closed form. Is it possible to give a solution for this which isn't a closed form solution? A.k.a expressible with special functions? – Kugelblitz Feb 26 '15 at 1:26
• Well technically contour integration also works on finite trigonometric integrals, they don't just have to be improper, for example the integral $$\int^{\pi}_{-\pi} \frac{1}{1 + 3(\cos{t})^2} dt$$ – Triatticus Mar 1 '15 at 2:56
• Well, in this case we have an $\ln$ not just trigometric. But contour integration won't work here since this is indefinite integral. Yeah , in this case you can calculate the trig. integral you pose using contour integration over the unit circle... but this does not hold in our case. Anyway, as I mentioned before contour integration does not apply here because we have an indefinite integral. – Tolaso Mar 1 '15 at 10:08

Simplifications such as limits as when R approaches $\infty$ or $0$ are not the only reasons to use contour integration. Sometimes properties of symmetry can give nice simplifications.

What may be useful in this case :

$\cos(a+bi) + \cos(a-bi)$ and $\sin(a+bi) + \sin(a-bi)$ are real

$\cos(a+bi) - \cos(a-bi)$ and $\sin(a+bi) - \sin(a-bi)$ are imaginary.

The reason to consider $a+bi$ and $a-bi$ is that we can easily build one upper and one lower half plane contour and investigate their sums and differences.