How to arrive at $4e^5-\frac{23}2$ from an exponential-like series?

Question

Evaluate the integral by changing to polar coordinates:

$$\iint_R ye^x\,\mathrm dA$$

where $R$ is the region in the first quadrant bounded by the $x$- and $y$-axes and the circle $x^2+y^2=25$.

I've converted to polar coordinates and found that the integral has the same value as

$$\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta\, e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta$$

Integrating by parts twice with respect to $r$ and substituting $\mu=5\cos\theta$, I've managed to reduce this to the single integral

$$25\int_{\mu=0}^{\mu=5}\left(\left(\frac1\mu-\frac2{\mu^2}+\frac2{\mu^3}\right)e^\mu-\frac2{\mu^3}\right)\,\mathrm d\mu$$

Now, this has an elementary antiderivative, but attempting to find it manually entails using the exponential integral function $\mbox{Ei}(x)$ in some of the intermediate steps. Lacking familiarity with the special function, I'm resorting to the power series for $e^x$ to compute this integral.

Expanding the integrand as needed, I get

$$25\int_{\mu=0}^{\mu=5}\left(\frac13+\frac\mu4+\frac{\mu^2}{10}+\frac{\mu^3}{36}+\frac{\mu^4}{168}+\frac{\mu^5}{960}+\cdots\right)\,\mathrm d\mu$$

$$=\frac{5^3}{1\times3}+\frac{5^4}{2\times4}+\dfrac{5^5}{3\times10}+\frac{5^6}{4\times36}+\frac{5^7}{5\times168}+\frac{5^8}{6\times960}+\cdots$$

The actual value of the integral is $4e^5-\dfrac{23}2$, and numerical comparisons suggest this summation gives the same value. I've intentionally left my result in its current form with the hopes of making it easier to see the next step, but I'm not getting it just yet.

How can I arrive at this value from the summation above?

Answer 1

Since $e^x = \sum_{n=1}^\infty x^n/n!$, we have $$ \frac1\mu e^\mu = \sum_{n=-1}^\infty \frac{1}{(n+1)!} \mu^n $$ $$ -\frac2{\mu^2} e^\mu = \sum_{n=-2}^\infty \frac{-2 }{(n+2)!}\mu^n $$ $$ \frac2{\mu^3} e^\mu = \sum_{n=-3}^\infty \frac{-2 }{(n+3)!}\mu^n $$ Adding these together along with the term $-2/\mu^3$, we get \begin{align} \left(\frac1\mu-\frac2{\mu^2}+\frac2{\mu^3}\right)e^\mu-\frac2{\mu^3} &= \sum_{n = 0}^\infty \left[ \frac{1}{(n+1)!} - \frac{2}{(n+2)!} + \frac{2}{(n+3)!} \right]\mu^n \\ &=\sum_{n = 0}^\infty \left[ \frac{(n+2)(n+3) - 2 (n+3) + 2}{(n+3)!} \right] \mu^n \\ &= \sum_{n = 0}^\infty \frac{(n+1)(n+2)}{(n+3)!} \mu^n \end{align} The antiderivative of this quantity with respect to $\mu$ is then \begin{align} F(\mu) = \int \left[ \left(\frac1\mu-\frac2{\mu^2}+\frac2{\mu^3}\right)e^\mu-\frac2{\mu^3} \right] d\mu &= \sum_{n=0}^\infty \frac{n+2}{(n+3)!} \mu^{n+1} \\ &= \sum_{n=0}^\infty \frac{(n+3) - 1}{(n+3)!} \mu^{n+1} \\ &= \sum_{n=0}^\infty \frac{1}{(n+2)!} \mu^{n+1} - \sum_{n=0}^\infty \frac{1}{(n+3)!} \mu^{n+1} \\ &= \sum_{n=2}^\infty \frac{1}{n!} \mu^{n-1} - \sum_{n=3}^\infty \frac{1}{n!} \mu^{n-2} \\ &= \frac{1}{\mu} \sum_{n=0}^\infty \frac{1}{n!} \mu^{n} - \frac{1}{\mu^2} \sum_{n=3}^\infty \frac{1}{n!} \mu^{n} \\ &= \frac{1}{\mu} \left[ e^\mu - (1 + \mu) \right] - \frac{1}{\mu^2} \left[ e^\mu - \left( 1 + \mu + \frac{\mu^2}{2!} \right) \right]\\ &= e^\mu \left[ \frac{1}{\mu} - \frac{1}{\mu^2} \right] - \frac{1}{2} + \frac{1}{\mu^2}. \end{align} Since $F(5) = \frac{4}{25} e^5 - \frac{23}{50}$ and $\lim_{\mu \to 0} F(\mu) = 0$, we conclude that the definite integral from $\mu = 0$ to $5$ is equal to $\frac{4}{25} e^5 - \frac{23}{50}$. Multiplying by the factor of 25 yields the desired result.

(Or you could reverse the order of integration and avoid all of the above, but where's the fun in that?)

Answer 2

It is worth to integrate with respect to $\theta$ for first: that leaves an elementary integral in $dr$,

$$ \int_{0}^{5}\int_{0}^{\pi/2}r^2 \sin(\theta)e^{r\cos\theta}\,d\theta\,dr = \int_{0}^{5}r^2\cdot\frac{e^r-1}{r}\,dr $$ namely: $$ \int_{0}^{5} r(e^r-1)\,dr = \left[(r-1)e^r-\frac{r^2}{2}\right]_{0}^{5} = 4e^5-\frac{23}{2}.$$