Maximum of a function defined via a definite integral exploring a problem I have introduced a function:
$$I(x)=\int_0^{\pi/2}xe^{-x\sin t} dt.$$
To my surprise the maximum of the function appears to be achieved at a value seemingly equal to $e$. I have checked it by computing the derivative of the function:
$$\frac{dI}{dx}=\int_0^{\pi/2}e^{-x\sin t}(1-x\sin t)\ dt$$
by an  online integrator and have got $x_{\rm max}\approx0.9996e$. I am not however aware of the possible integration errors of the routine.
Could you please check it with a  tool providing not only the result of the integration but also its error? Of course an analytical proof of the equality or inequality would be even more appreciated.
 A: We have:
$$ I(x)=\int_{0}^{\pi/2}x\,e^{-x\sin t}\,dt = \frac{\pi x}{2}\left(I_0(x)-L_0(x)\right) $$
where $I_0$ and $L_0$ are a Bessel and a Struve function. The fact that a zero of
$$ I'(x) = \left(I_0(x)-L_0(x)\right) + x\left(I_1(x)-L_{-1}(x)\right) $$
occurs very close to $e$ probably depends on the continued fraction expansion of $I_0$ and $L_0$. However, $e$ is not an exact zero:
$$ I'(e) \approx -0.0000519.$$
A: The integral has the closed form (given by Maple)
$$I(x)=\int_0^{\pi/2}x e^{-x \sin t}dt=\frac{\pi}{2}x \left( I_0(x) -\mathbf{L_0}(x)\right)$$
with the modified Bessel and Struve functions, see http://dlmf.nist.gov/10.25 and http://dlmf.nist.gov/11.2. With this closed form you can solve $I'(x)=0$ and get the maximum at $x_m=2.71729791967257323761$ with $x_m-e\approx -0.0009839$
A: 
$$x_\max\simeq2.717297919672573237610594072000\ldots$$

Also, the integral evaluates to $~\dfrac\pi2~x~\Big[I_0(x)-L_0(x)\Big].~$ See Bessel and Struve functions for more information.
A: $$I(x)=\int_0^{\pi/2}xe^{-x\sin t} dt=\frac{1}{2} \pi  x (I_0(x)-\pmb{L}_0(x))$$ where appear Bessel and Struve functions. For the derivative $$I'(x)=\frac{1}{2} \pi  (-x \pmb{L}_{-1}(x)-\pmb{L}_0(x)+I_0(x)+x I_1(x))$$ it cancels for $x\approx 2.71729791967200$ which is very close to $e$ but not identical.
