Integral of exponential over square root

I'm trying to integrate

$$\int_0^{x

I've tried various substitutions to no avail. Any ideas?

• Is $a$ a constant? – Brian Mar 21 at 22:50
• Yes, sorry! I'll edit the question. – SabrinaChoice Mar 21 at 22:53
• Your system has a singularity when $x=a$, so the interval of integration needs to be specified or else the integral will inevitably blow up. – aghostinthefigures Mar 21 at 22:54
• Thanks, you're right. I'll edit as such. – SabrinaChoice Mar 21 at 22:55
• @DanielMcLaury I don't have any reason to believe it can be written in closed form. Just hoping it's possible. – SabrinaChoice Mar 21 at 23:31

To expand on my comment, trying the $$a = 1$$ and $$a = 2$$ cases in WolframAlpha gives

$$\int_0^1 \frac{e^x}{\sqrt{x(1-x)}}\, dx = \sqrt{e} \, \pi \, I_0(1/2)$$

and

$$\int_0^2 \frac{e^x}{\sqrt{x(2-x)}}\, dx = e \, \pi \, I_0(1)$$

suggesting that the general solution might be

$$\int_0^a \frac{e^x}{\sqrt{x(a-x)}}\, dx = e^{a/2} \pi I_0(a/2).$$

where $$I_0$$ is a modified Bessel function of the first kind.

UPDATE:

Now that I have a second, I was able to do the substitutions suggested by the form of the expressions above and it indeed works out:

$$\int_0^a \frac{e^x \, dx}{\sqrt{x(a-x)}} = \int_0^{2b} \frac{e^x \, dx}{\sqrt{x(2b-x)}}$$

(where $$a=2b$$),

$$= \int_0^{2b} \frac{e^x \, dx}{\sqrt{b^2 - (x-b)^2}} = \int_{-1}^{1} \frac{b e^{b u + b} \, du}{\sqrt{b^2 - (b u)^2}}$$

(where $$b u = x + b$$),

$$= e^b \int_{-1}^{1} \frac{e^{b u} \, du}{\sqrt{1 - u^2}} = e^b \int_{-\pi/2}^{\pi/2} \frac{e^{b \sin\theta} \, \cos \theta \, d\theta}{\sqrt{1 - \sin^2 \theta}}$$

(where $$u = \sin \theta$$)

$$= e^b \int_{-\pi/2}^{\pi/2} e^{b \sin\theta} d\theta = e^b \pi I_0(b) = e^{a/2} \pi I_0\left(\frac{a}{2}\right)$$

Here $$I_0(x) = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} e^{x \sin\theta} \, d\theta$$

• This is for sure a very nice solution for the particular case of $\int_0^a \frac{e^x \, dx}{\sqrt{x(a-x)}}$ . We still the problem with $\int_0^t \frac{e^x \, dx}{\sqrt{x(a-x)}}$ when $t < a$. Then $\to +1$. Cheers. – Claude Leibovici Mar 24 at 2:57
• Perhaps I didn't understand the question, then. I thought x was the variable of integration. If we have an arbitrary upper bound then as you say the problem is harder. – Daniel McLaury Mar 24 at 3:17
• For this particular case $t=a$, it is an interesting solution. May I confess that I tried even the case $t=a$ using several CAS with no success at all. – Claude Leibovici Mar 24 at 3:22

I am not sure that any closed form could be found and, may be, only series expansions could be the way to go.

First, let $$x=ay$$ to make $$I=\int_0^{t} {\frac{e^x}{\sqrt{x(a-x)}}\, dx}=\int_0^{\frac ta} {\frac{e^{ay}}{\sqrt{y(1-y)}}\, dy}$$ Expanding around $$y=0$$ would give $$\frac{1}{\sqrt{y(1-y) }}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}y^{n-\frac 12}$$ and $$\int e^{ay}\,y^{n-\frac 12}\,dy=-y^{n+\frac{1}{2}} E_{\frac{1}{2}-n}(-a y)$$ where appears the exponential integral function.