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#1 h Apr 24, 2025, 8:43 PM

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Determine the value of $\int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, \mathrm dx .$

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#2 h Apr 25, 2025, 12:08 AM • 2 Y

Y by ap246, tom-nowy

From integration by parts with $u'=e^x$ and $v=\sin\left(\sqrt{1-x^2}\right)$ , we get
$\begin{align*} \int_{-1}^1 e^x\sin\left(\sqrt{1-x^2}\right)\,dx&=e^x\sin\left(\sqrt{1-x^2}\right)\Bigg|_{-1}^1-\int_{-1}^1 e^x\cdot \cos\left(\sqrt{1-x^2}\right)\cdot \frac{-x}{\sqrt{1-x^2}}\,dx\\ &=\int_{-1}^1 e^x\cos\left(\sqrt{1-x^2}\right)\cdot\frac{x}{\sqrt{1-x^2}}\,dx. \end{align*}$ Now substitute $x=\cos\theta$ , and we get $\int_0^\pi e^{\cos\theta}\cos(\sin\theta)\cos\theta\,d\theta=\frac{1}{2}\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)\cos\theta\,d\theta.$ Observe that $e^{e^{i\theta}}=e^{\cos\theta+i\sin\theta}=e^{\cos\theta}\left(\cos(\sin\theta)+i\sin(\sin\theta)\right),$ so if we let $z=e^{i\theta}$ , we get
$\begin{align*} e^{\cos\theta}\cos(\sin\theta)\cos\theta&=\text{Re}(e^{e^{i\theta}})\cdot\frac{e^{i\theta}+e^{-i\theta}}{2}\\ &=\text{Re}(e^z)\cdot \frac{z+z^{-1}}{2}\\ &=\frac{e^z+e^{1/z}}{2}\cdot\frac{z+z^{-1}}{2}. \end{align*}$ Then $dz=izd\theta$ , and it becomes a contour integral around the unit circle:
$\begin{align*}&\,\,\,\,\,\,\frac{1}{2}\int_{|z|=1} \frac{1}{4}\left(e^z+e^{1/z}\right)\left(z+\frac{1}{z}\right)\cdot\frac{1}{iz}\,dz\\ &=\frac{1}{8i}\int_{|z|=1} \left(e^z+e^{1/z}\right)\left(1+\frac{1}{z^2}\right)\,dz. \end{align*}$ Finally, we apply the residue theorem. We get a residue of $1$ from $\dfrac{e^z}{z^2}$ and a residue of $1$ from $e^{1/z}$ , so the final value is $\frac{1}{8i}\cdot 2\pi i\cdot 2=\boxed{\frac{\pi}{2}}.$

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#3 h Apr 25, 2025, 12:01 PM • 1 Y

Y by tom-nowy

Given positive integer $n,$ let
$I_n:=\int_{-1}^1 e^x\sin(\sqrt{1-x^2})\sin(n \arccos x)\frac{dx}{\sqrt{1-x^2}}.$ Notice that $I_1$ is your integral.

Claim. $I_n=\frac{\pi}{2n!}.$

Proof. Put $x=\cos t.$ Then
$I_n=\int_0^{\pi}e^{\cos t}\sin(\sin t)\sin(nt) \ dt=\frac{1}{2i}\int_0^{\pi}e^{\cos t}(e^{i\sin t}-e^{-i\sin t})\sin(nt) \ dt$ $=\frac{1}{2i} \int_0^{\pi}\left(e^{e^{it}}-e^{e^{-it}}\right)\sin(nt) \ dt=\frac{1}{2i}\int_0^{\pi}\sum_{k \ge 0}\frac{e^{ikt}-e^{-ikt}}{k!} \sin(nt) \ dt$ $=\int_0^{\pi}\sum_{k \ge 0}\frac{\sin(kt)}{k!}\sin(nt) \ dt=\sum_{k \ge 0}\frac{1}{k!}\int_0^{\pi}\sin(kt)\sin(nt) \ dt=\frac{1}{n!}\int_0^{\pi}\sin^2(nt) \ dt=\frac{\pi}{2n!}.$

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#4 h Apr 25, 2025, 2:43 PM • 1 Y

Y by tom-nowy

Here's another generalization of your problem:
$\int_{-1}^1\frac{e^x\sin(\sqrt{1-x^2})}{t^2-2xt+1} \ dx=\frac{\pi(e^t-1)}{2t}, \ \ \ \ \ \ t \in (-1,1), \ \ \ \ \ \ \ \ (*)$ where at $t=0$ the RHS is evaluated as its limit at $t=0,$ which is $\pi/2.$
That follows from my previous post; just look at the power series $\sum_{n \ge 1}I_nt^n$ and use the following identity
$\sum_{n \ge 1}\sin(n\theta)t^n=\frac{t\sin \theta}{t^2-2t\cos \theta +1}.$
PS. Taking limit as $t \to \pm 1,$ we see that $(*)$ also holds for $t=\pm 1.$ For $|t| > 1,$ changing $t$ to $1/t$ in $(*)$ gives
$\int_{-1}^1\frac{e^x\sin(\sqrt{1-x^2})}{t^2-2xt+1} \ dx=\frac{\pi(e^{1/t}-1)}{2t}, \ \ \ \ \ \ |t| > 1.$

This post has been edited 4 times. Last edited by ysharifi, Apr 26, 2025, 7:07 PM
Reason: PS added

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