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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.

user
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4 Answers4

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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: $$ I'(e) \approx -0.0000519.$$

Jack D'Aurizio
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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 $x_m-e\approx -0.0009839$

gammatester
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$$x_\max\simeq2.717297919672573237610594072000\ldots$$

Also, the integral evaluates to $~\dfrac\pi2~x~\Big[I_0(x)-L_0(x)\Big].~$ See Bessel and Struve functions for more information.

Community
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Lucian
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$$I(x)=\int_0^{\pi/2}xe^{-x\sin t} dt=\frac{1}{2} \pi x (I_0(x)-\pmb{L}_0(x))$$ 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.

Claude Leibovici
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