# Integral $\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$

$$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$

Thanks!

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Maybe $x=\cos(t)$ helps. – draks ... Apr 9 '12 at 18:57
Hint: $4.0$ – Fabian Apr 9 '12 at 18:59
To make use of draks' hint, you'll need to use the integral representation of the modified Bessel functions of the first kind. – joriki Apr 9 '12 at 19:03
$\pi I_0(1) \approx 3.977 \ne 4.0$ – GEdgar Apr 9 '12 at 20:30

$$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$ Substitution:$x=\sin\theta$

$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta$$

$$e^{\sin\theta}=1+\frac{\sin\theta}{1!}+\frac{\sin^2\theta}{2!}+\frac{\sin^3\theta}{3!}.......$$ The odd powers of sine when used in the integral will produce zero. Again for even values of n we have $$\int_{-\pi/2}^{\pi/2}\sin^n\theta d\theta=2 \times \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}$$ Therefore,

$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta=\pi+2\times\Sigma\frac{1}{n!}[\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}]$$

$$=\pi+\pi\Sigma [\frac{1}{n(n-2)(n-4)......2}]^2$$ ["n" running from two through even values ]

Writing n=2m we have,

$$\pi<I=\pi\Sigma[\frac{1}{2^m}\frac{1}{m!}]^2<4$$

[m running from 0 to infinity]

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Let me try to sort out the previous takes. For this, we will have to take as a prerequisite the series for the zeroth-order modified Bessel function of the first kind

$$I_0(x)=\sum_{k=0}^\infty \frac1{(k!)^2}\left(\frac{x^2}{4}\right)^k$$

which can be derived from the modified Bessel differential equation

$$x^2 y^{\prime\prime}+xy^\prime=x^2 y$$

with initial conditions $y(0)=1,y^\prime(0)=0$.

Now,

\begin{align*} \int_{-1}^1 \frac{\exp\,x}{\sqrt{1-x^2}}\mathrm dx&=\sum_{k=0}^\infty \frac1{k!}\int_{-1}^1 \frac{x^k}{\sqrt{1-x^2}}\mathrm dx\\ &=\sum_{k=0}^\infty \frac1{(2k)!}\int_{-1}^1 \frac{(x^2)^k}{\sqrt{1-x^2}}\mathrm dx\\ &=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^1 \frac{(x^2)^k}{\sqrt{1-x^2}}\mathrm dx\\ &=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^{\pi/2} \frac{\sin^{2k} u \; \cos\,u}{\sqrt{1-\sin^2 u}}\mathrm du\\ &=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^{\pi/2} \sin^{2k} u\mathrm du\\ &=\pi\sum_{k=0}^\infty \frac1{(2k)!}\frac{(2k)!}{4^k (k!)^2}\\ &=\pi\sum_{k=0}^\infty \frac1{(k!)^2}\left(\frac14\right)^k=\pi I_0(1)\\ \end{align*}

where we used the odd/even properties of the moments in the second and third lines, and the Wallis formula in the fifth line.

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