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

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


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.



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.)

  • $\begingroup$ 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. $\endgroup$ – user557032 Apr 28 '18 at 17:45
  • $\begingroup$ You may have forgot a $k$-th power on the $2$ outside of the finite sum. $\endgroup$ – giobrach Apr 28 '18 at 17:49
  • $\begingroup$ yes ;;;;;;;;;;;;;;; $\endgroup$ – 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.


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