Let $H:\mathbb{R} \to \mathbb{R}$ be the function $H(x)=\frac{e^x+e^{-x}}{2} $ for $ x \in \mathbb{R}$ Let $H:\mathbb{R} \to \mathbb{R}$ be the function $H(x)=\dfrac{e^x+e^{-x}}{2} $ for $ x \in \mathbb{R}$
Let $\displaystyle f(x) = \int_{0}^{\pi} H(x\sin(\theta))d\theta$
Then is the following equation correct?
$xf''(x)-f'(x)+xf(x)=0$
The first thing that comes to my mind is to use the Leibnitz rule. However, Leibnitz rule is generally used for equations in this form
$\displaystyle g(x)= \int_{f_1(x)}^{f_2(x)}h(t)dt$
So I thought of substituting $\sin(\theta) = t$. Then I get something of this form
$\displaystyle f(x)= 2\int_{0}^{1} \frac{H(xt)}{\sqrt{1-t^2}}dt$
How do I proceed? Also, I need some insights on why should I proceed in a particular direction.
 A: Hint
Since $H(x)=\cosh(x)$, you want to compute
$$ f(x) = \int_{0}^{\pi} \cosh (x \sin (\theta ))\,d\theta$$ Using the expansion of the hyperbolic cosine
$$f(x)=\sum_{n=0}^\infty  \int_{0}^{\pi} \frac{(x \sin (\theta ))^{2 n}}{(2 n)!}\,d\theta=\sum_{n=0}^\infty \frac{x^{2n} }{(2 n)!} \int_{0}^{\pi} \sin^{2n} (\theta)\,d\theta $$
$$f(x)=\sum_{n=0}^\infty \frac{x^{2n} }{(2 n)!} \frac{\sqrt{\pi }\,\, \Gamma \left(n+\frac{1}{2}\right)}{\Gamma
   (n+1)}=\pi\sum_{n=0}^\infty  \frac{ 1}{(2 n)!} \frac{ \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\,\,\Gamma
   (n+1)}\,x^{2n}=\pi \sum_{n=0}^\infty \frac{1}{(n!)^2}\left(\frac{x}{2}\right)^{2 n}$$ and the summation corresponds to a well known function.
A: The Leibniz rule has various forms depending on the scenario. You want:
$$\frac d{dx}\int_a^b f(x,t)dt=\int_a^b\frac{\partial}{\partial x}f(x,t)dt$$
so in your case:
$$f'(x)=\int_0^1\frac{\partial H(xt)}{\partial x}\frac{1}{\sqrt{1-t^2}}dt$$
$$f''(x)=\int_0^1\frac{\partial^2 H(xt)}{\partial x^2}\frac{1}{\sqrt{1-t^2}}dt$$
you can also group your functions together into one integral to give:
$$\int_0^1 \left[x\frac{\partial^2 H(xt)}{\partial x^2}-\frac{\partial H(xt)}{\partial x}+xH(xt)\right]\frac{1}{\sqrt{1-t^2}}dt\equiv0$$
which is now what you want to prove. To simplify matters:
$$\frac{\partial^2 H(xt)}{\partial x^2}=t^2H''(xt)$$
$$\frac{\partial H(xt)}{\partial x}=tH'(xt)$$
This should be enough information for you to continue
