I have again a doubt regarding an exercise of differentiation under the integral sign. In this case, it concerns the integral:

$$ \int_0^{2π} e^{\cos(x)}\cos(\sin x) \, \mathrm{d}x $$

I tried the substitution:

$f(a,b) = \int_0^{2π} e^{acos(x)}\cos(b\sin x) \, \mathrm{d}x $

So then:

$\frac {\partial f} {\partial a} = \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x $


$\frac {\partial f} {\partial b} = -\int_0^{2π} e^{a\cos(x)}\sin(b\sin x)\sin x \, \mathrm{d}x $

Then what i did was try to integrate $\frac{\partial f}{\partial b}$ by parts, making:

$dv = -e^{a\cos x}\sin x\, \mathrm{d}x$


$v = \frac{1}{a}e^{a\cos x}$


$u = \sin(b\sin x)$


$du = \cos(b\sin x)b\cos x$

Then we got that:

$\frac{\partial f}{\partial b} = - \frac{b}{a} \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x = -\frac{b}{a} \frac{\partial f}{\partial a} $

(Because evaluating $uv$ from $0$ to $2π$ gives us $0$)

So we get the partial differential equation:

$\frac{1}{b} \frac{\partial f}{\partial b} = - \frac{1}{a} \frac{\partial f}{\partial a}$

I solved this by separation of variables, making $f(a,b) = A(a)B(b)$. The end result was:

$f(a,b) = Ce^{\frac{h^2}{2}(a^2 - b^2)}$

Where $C$ is an integration constant and $h$ comes from solving the ODE associated with each $A$ and $B$.

So, my problem begins with this, because I am not sure of my result, this is mostly because of the $h$, should I pick an specific value? Or my result is just wrong? If this is the case, where did I made the mistake? Because evaluating $f(0,0)$ we get: $f(0,0) = C = 2π$, and so: $f(1,1) = 2π$, that is the correct result evaluating in wolfram alfa. I just want to know if there is a way in wich I should get the value.


Let $$I(a)=\int_{0}^{2\pi}e^{a\cos x}\cos(a\sin x)dx $$ Then \begin{align} I’(a) &=\int_0^{2\pi} e^{a\cos x}(\cos(a\sin x)\cos x- \sin(a\sin x)\sin x)dx\\ &=\frac1a\int_{0}^{2\pi}d(e^{a\cos x}\sin(a\sin x))=0\\ \end{align} which leads to $$\int_{0}^{2\pi}e^{\cos x}\cos(\sin x)dx = I(1) = I(0)=2\pi$$


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