# Solving $\int_{0}^{\frac{\pi}{2}} e^{a \cos(x)} dx$

$$\displaystyle\int_{0}^{\frac{\pi}{2}} e^{a \cos(x)}\,\mathrm{d}x$$

I tried this:

$$=\displaystyle\int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty}\frac{a^k \cos^k(x)}{k!}\,\mathrm{d}x$$

Since $\frac{\cos^k(x)}{k!} \leq \frac{1}{k!}$, then the series converges uniformly by Weierstrass test

$$=\displaystyle\sum_{k=0}^{\infty}\frac{a^k}{k!}\int_{0}^{\frac{\pi}{2}}\cos^k(x)\,\mathrm{d}x$$

But then the problem is this last integral. Wolfram's answer is pretty nice, then it's possible there's an easy way to solve it that I'm not seeing.

Thanks

The integral $$I_k:=\int_0^{\pi/2} \cos(x)^k \mathrm{d} x$$ can be expressed in a closed formula. For $$k>1$$ wet by partial integration the following recurrence: We have $$\int_0^{\pi/2} \cos(x)^k \mathrm{d} x = (k-1)\int_0^{\pi/2} \sin(x)^2 \cos(x)^{k-2} \mathrm{d} x$$ and taking in mind that $$\sin(x)^2 = 1 - \cos(x)^2$$ we get $$\int_0^{\pi/2} \cos(x)^k \mathrm{d} x = \frac{k-1}{k} \int_0^{\pi/2} \cos(x)^{k-2} \mathrm{d} x.$$ Thus, we get for odd $$k$$ $$\tag{1}I_k = \prod_{i=1}^{(k-1)/2} \frac{2i}{2i+1} = 2^{k-1} \frac{(((k-1)/2)!)^2}{k!}$$ and for even $$k$$ $$\tag{2}I_k = \frac{\pi}{2} \prod_{i=1}^{k/2} \frac{2i-1}{2i} = \frac{\pi}{2^{k+1}} \frac{k!}{((k/2)!)^2}.$$ This can be rewritten in terms of the gamma-function as follows. $$I_k = \frac{\sqrt{\pi}}{2} \frac{\Gamma(k/2+1/2)}{\Gamma(k/2+1)}$$ Next we split the initial sum according to even or odd $$k$$: Using (2) we get for even $$k$$ the subsum $$\tag{3}\sum_{k=1}^\infty \frac{a^{2k}}{(2k)!} I_{2k} = \frac{\pi}{2} \sum_{k=1}^\infty \frac{a^{2k}}{4^{k} (k!)^2}$$ and for odd $$k$$ the subsum $$\tag{4}\sum_{k=1}^\infty \frac{a^{2k-1}}{(2k-1)!} I_{2k-1} = \sum_{k=1}^\infty a^{2k-1} 4^{k-1} \Big(\frac{(k-1)!}{(2k-1)!}\Big)^2.$$ Maple says that (4) is equal to $$\frac{\pi}{2}\mathrm{StruveL}(0,a)$$ and (3) is equal to $$\frac{\pi}{2} \Big( \mathrm{BesselI}(0,a)-1 \Big).$$ This can be checked by comparing the series representations of these two functions.

I think you can use the identity $$\cos x= \frac{e^{ix} + e^{-ix}}{2}$$ and then the binomial theorem. Then you’d be integrating exponentials, which is relatively easy.

(Of course after that you’d need to deal with the infinite sum of a variable-upper-bound finite sum, but then again maybe Taylor series wasn’t the quickest approach to this question – unless there are more useful trig identities that I’m not remembering.)

• Thanks. I tried your suggestion and got $\sum_{k=0}^{\infty}\frac{a^k}{2k!}\sum_{j=0}^{k}\binom{k}{j}\frac{e^{(2ij-ik)\frac{\pi}{2}}-1}{2ij-ik}$, but I don't see how Gamma functions will appear from that. – user557032 Apr 28 '18 at 17:45
• You may have forgot a $k$-th power on the $2$ outside of the finite sum. – giobrach Apr 28 '18 at 17:49
• yes ;;;;;;;;;;;;;;; – user557032 Apr 28 '18 at 17:50

The integral $\int_0^{\pi/2}\cos^k x dx$ can be computed as follows.

By definition of $\Gamma$-function: $$\Gamma(q)=\int_0^\infty t^{p-1}e^{-t}dt\stackrel{t=x^2}=2\int_0^\infty x^{2p-1}e^{-x^2}dx$$ Then $$\begin {eqnarray} \Gamma(p)\Gamma(q)=&4\int_0^\infty x^{2p-1}e^{-x^2}dx\int_0^\infty y^{2q-1}e^{-y^2}dy\\ =&4\iint x^{2p-1} y^{2q-1}e^{-x^2-y^2}dxdy\\ =&4\int_0^\infty r^{2p+2q-1} e^{-r^2}dr\int_0^{\pi/2}\cos^{2p-1}\phi \sin^{2q-1}\phi\;d\phi\\ =&2\Gamma(p+q)\int_0^{\pi/2}\cos^{2p-1}\phi \sin^{2q-1}\phi\;d\phi\\ \end {eqnarray}\\ \Rightarrow \int_0^{\pi/2}\cos^{2p-1}\phi \sin^{2q-1}\phi\;d\phi=\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}.$$

Substituting in this formula $p=\frac{k+1}{2}$, $q=\frac{1}{2}$ gives the result.