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Recently, I saw these two calculus problems showed up on internet. I tried to answer those problems but I couldn't. I've already used WolframAlpha to figure it out, but the result was nothing. These are the problems: $$ \int\left(\pi^{\sin^2x}+e^{\sin^2x}\right)^2\cos2x~dx $$ and $$ \int\frac{\sin x}{1+\cos x+e^x}dx $$ Could anyone here help me out how to answer these problems? I'd be grateful for any help you are able to provide.

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Your two integrals don't have any closed forms according to Mathematica, in other means they can't be written in terms of the usual functions such as $\ln, a^x, \sin,\ldots$

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    $\begingroup$ Thanks Hakim. These two integrals have been bugging me for very long time and I still try to figure them out by considering them as definite integrals. $\endgroup$ – Tunk-Fey Aug 4 '14 at 18:27
  • $\begingroup$ You're welcome @Tunk-Fey ;) $\endgroup$ – Hakim Aug 5 '14 at 10:44
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For $\int\left(\pi^{\sin^2x}+e^{\sin^2x}\right)^2\cos2x~\text{d}x$ ,

$$\begin{align} \int\left(\pi^{\sin^2x}+e^{\sin^2x}\right)^2\cos2x\ \text{d}x & = \int\left(\pi^{2\sin^2x}+2\pi^{\sin^2x}e^{\sin^2x}+e^{2\sin^2x}\right)(1-2\sin^2x)\ \text{d}x \\ & = \int\left(e^{2\ln\pi\sin^2x}+2e^{\ln\pi\sin^2x}e^{\sin^2x}+e^{2\sin^2x}\right)(1-2\sin^2x)\ \text{d}x \\ & = \int\left(e^{2\ln\pi\sin^2x}+2e^{(\ln\pi+1)\sin^2x}+e^{2\sin^2x}\right)(1-2\sin^2x)\ \text{d}x \\ \end{align} $$

All terms are either in the form $\displaystyle \int e^{a\sin^2x}\ \text{d}x$ or $\displaystyle \int e^{a\sin^2x}\sin^2x\ \text{d}x$

For $\displaystyle \int e^{a\sin^2x}\ \text{d}x$ ,

$$ \int e^{a\sin^2x}\ \text{d}x=\int\sum\limits_{n=0}^\infty\dfrac{a^n\sin^{2n}x}{n!}\ \text{d}x=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{a^n\sin^{2n}x}{n!}\right)\ \text{d}x $$

For $n \in \Bbb{N}$, we have

$$\int\sin^{2n}x\ \text{d}x=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$$

Therefore, $$ \begin{align} \int\left(1+\sum\limits_{n=1}^\infty\dfrac{a^n\sin^{2n}x}{(2n)!}\right)dx & = x+\sum\limits_{n=1}^\infty\dfrac{a^n(2n)!x}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^n(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^3(2k-1)!}+C \\ & = \sum\limits_{n=0}^\infty\dfrac{a^n(2n)!x}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{a^n(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^3(2k-1)!}+C \end{align} $$

For $\displaystyle\int e^{a\sin^2x}\sin^2x\ \text{d}x$ ,

$$ \int e^{a\sin^2x}\sin^2x\ \text{d}x=\int\sum\limits_{n=0}^\infty\dfrac{a^n\sin^{2n+2}x}{n!}\text{d}x $$

For $n$ is any non-negative integer,

$$ \int\sin^{2n+2}x\ \text{d}x = \dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C $$

Thus

$$\int\sum\limits_{n=0}^\infty\dfrac{a^n\sin^{2n+2}x}{n!}\ \text{d}x=\sum\limits_{n=0}^\infty\dfrac{a^n(2n+1)!x}{2^{2n+1}(n!)^2(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^n(2n+1)!(k!)^2\sin^{2k+1}x\cos x}{2^{2n-2k+1}(n!)^2(n+1)!(2k+1)!}+C$$

For $\int\dfrac{\sin x}{1+\cos x+e^x}~\text{d}x$ ,

$$\begin{align} \int\dfrac{\sin x}{1+\cos x+e^x}~\text{d}x & =\int\dfrac{e^{-x}\sin x}{e^{-x}(1+\cos x)+1}~\text{d}x \\ & =\int\sum\limits_{n=0}^\infty(-1)^ne^{-(n+1)x}(1+\cos x)^n\sin x~\text{d}x \\ & =-\int\sum\limits_{n=0}^\infty(-1)^ne^{-(n+1)x}(1+\cos x)^n~\text{d}(1+\cos x) \\ & =-\int\sum\limits_{n=0}^\infty(-1)^ne^{-(n+1)x}~\text{d}\left(\dfrac{(1+\cos x)^{n+1}}{n+1}\right) \\ & =\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}e^{-(n+1)x}(1+\cos x)^{n+1}}{n+1}+\int\left(\dfrac{(1+\cos x)^{n+1}}{n+1}\right)~\text{d}\left(\sum\limits_{n=0}^\infty(-1)^ne^{-(n+1)x}\right) \\ & =-\ln(1+e^{-x}(1+\cos x))+\int\sum\limits_{n=0}^\infty(-1)^{n+1}e^{-(n+1)x}(1+\cos x)^{n+1}~\text{d}x \\ & =-\ln(1+e^{-x}(1+\cos x))+\int\sum\limits_{n=0}^\infty(-1)^{n+1}e^{-(n+1)x}~\text{d}x+\int\sum\limits_{n=0}^\infty\sum\limits_{m=0}^nC_m^{n+1}(-1)^{n+1}e^{-(n+1)x}\cos^{n-m+1}x~\text{d}x \\ & =-\ln(1+e^{-x}(1+\cos x))-\int\dfrac{e^{-x}}{1+e^{-x}}~\text{d}x+\int\sum\limits_{m=0}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^{n+1}(n+1)!e^{-(n+1)x}\cos^{n-m+1}x}{m!(n-m+1)!}~\text{d}x \\ & =\ln(1+e^{-x})-\ln(1+e^{-x}(1+\cos x))+\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n+1}(m+n+1)!e^{-(m+n+1)x}\cos^{n+1}x}{m!(n+1)!}~\text{d}x \\ & =\ln\dfrac{1+e^{-x}}{1+e^{-x}(1+\cos x)}+\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+2n+1}(m+2n+1)!e^{-(m+2n+1)x}\cos^{2n+1}x}{m!(2n+1)!}~\text{d}x+\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+2n+2}(m+2n+2)!e^{-(m+2n+2)x}\cos^{2n+2}x}{m!(2n+2)!}~\text{d}x \\ & =\ln\dfrac{e^x+1}{e^x+\cos x+1}-\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^m(m+2n+1)!C_{n+k+1}^{2n+1}e^{-(m+2n+1)x}\cos((2k+1)x)}{4^nm!(2n+1)!}~\text{d}x+\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m(m+2n+2)!e^{-(m+2n+2)x}C_{n+1}^{2n+2}}{4^{n+1}m!(2n+2)!}~\text{d}x+\int\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^m(m+2n+2)!C_{n+k+2}^{2n+2}e^{-(m+2n+2)x}\cos((2k+2)x)}{2^{2n+1}m!(2n+2)!}~\text{d}x \\ & =\ln\dfrac{e^x+1}{e^x+\cos x+1}-\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m(m+2n+2)!e^{-(m+2n+2)x}}{4^{n+1}m!((n+1)!)^2(m+2n+2)}-\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^m(m+2n+1)!e^{-(m+2n+1)x}((2k+1)\sin((2k+1)x)-(m+2n+1)\cos((2k+1)x))}{4^nm!(n+k+1)!(n-k)!((m+2n+1)^2+(2k+1)^2)}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^m(m+2n+2)!e^{-(m+2n+2)x}((2k+2)\sin((2k+2)x)-(m+2n+2)\cos((2k+2)x))}{2^{2n+1}m!(n+k+2)!(n-k)!((m+2n+2)^2+(2k+2)^2)}+C \\ & =\ln\dfrac{e^x+1}{e^x+\cos x+1}-\sum\limits_{m=0}^\infty\sum\limits_{n=1}^\infty\dfrac{(-1)^m(m+2n)!e^{-(m+2n)x}}{4^nm!(n!)^2(m+2n)}-\sum\limits_{m=0}^\infty\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^m(m+2n+1)!e^{-(m+2n+1)x}((2k+1)\sin((2k+1)x)-(m+2n+1)\cos((2k+1)x))}{4^nm!(n+k+1)!(n-k)!((m+2n+1)^2+(2k+1)^2)}+\sum\limits_{m=0}^\infty\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^m(m+2n+2)!e^{-(m+2n+2)x}((2k+2)\sin((2k+2)x)-(m+2n+2)\cos((2k+2)x))}{2^{2n+1}m!(n+k+2)!(n-k)!((m+2n+2)^2+(2k+2)^2)}+C \\ & =\ln\dfrac{e^x+1}{e^x+\cos x+1}-\sum\limits_{m=0}^\infty\sum\limits_{n=1}^\infty\dfrac{(2m+2n)!e^{-(2m+2n)x}}{4^n(2m)!(n!)^2(2m+2n)}+\sum\limits_{m=0}^\infty\sum\limits_{n=1}^\infty\dfrac{(2m+2n+1)!e^{-(2m+2n+1)x}}{4^n(2m+1)!(n!)^2(2m+2n+1)}-\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^m(m+2n+2k+1)!e^{-(m+2n+2k+1)x}((2k+1)\sin((2k+1)x)-(m+2n+2k+1)\cos((2k+1)x))}{2^{2n+2k}m!n!(n+2k+1)!((m+2n+2k+1)^2+(2k+1)^2)}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^m(m+2n+2k+2)!e^{-(m+2n+2k+2)x}((2k+2)\sin((2k+2)x)-(m+2n+2k+2)\cos((2k+2)x))}{2^{2n+2k+1}m!n!(n+2k+2)!((m+2n+2k+2)^2+(2k+2)^2)}+C \\ & =\ln\dfrac{e^x+1}{e^x+\cos x+1}-\sum\limits_{m=0}^\infty\sum\limits_{n=1}^\infty\dfrac{(2m+2n)!e^{-(2m+2n)x}}{4^n(2m)!(n!)^2(2m+2n)}+\sum\limits_{m=0}^\infty\sum\limits_{n=1}^\infty\dfrac{(2m+2n+1)!e^{-(2m+2n+1)x}}{4^n(2m+1)!(n!)^2(2m+2n+1)}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{m+k}(m+2n+k)!e^{-(m+2n+k)x}(k\sin kx-(m+2n+k)\cos kx)}{2^{2n+k-1}m!n!(n+k)!((m+2n+k)^2+k^2)}+C \\ \end{align}$$

Which relates to Srivastava-Daoust Function

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    $\begingroup$ you really need to learn to use LaTex syntax properly. All of your posts contain these errors. Please use double dollar signs when doing integrals and sums, not $....$. We would be able to read your integral signs that way since they would be larger. $\endgroup$ – Jeff Faraci Apr 11 '14 at 23:32
  • $\begingroup$ This just seems to complicate things... (but did not down vote btw) $\endgroup$ – Winther May 12 '14 at 11:18
  • $\begingroup$ The two sums don't have any closed forms according to Mathematica. $\endgroup$ – Hakim Aug 4 '14 at 16:13

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