How to evaluate these integrals: $$\int_{-1}^{1}(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5xdx$$ $$\int_{-1}^{1}\frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x}{\sqrt{1-x^{2}}}dx$$ Can anybody help me ? Thank you!
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4It seems complicated. I wish $\sin 5x$ is $\cos 5x$. – Paul Dec 26 '11 at 12:00
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1Wolframalpha gives $0.295767$ for the first integral. And $1.36153$ for the second. – Raskolnikov Dec 26 '11 at 12:25
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Is this a homework or research problem? – Norbert Dec 26 '11 at 12:33
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@Paul, it is $sin5x$. It is a homework. What method can be used for solving this integral? – Roin Dec 26 '11 at 12:53
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1Roin: What similar cases do you know how to solve? Which techniques do you know which you tried and failed? What indications did you receive from your teacher? – Did Dec 26 '11 at 15:04
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@DidierPiau, solving of this integral is required for my given problem. – Roin Dec 26 '11 at 18:13
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2Roin: Here is what I call a zero-content comment. – Did Dec 26 '11 at 18:22
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3@Roin: The integrals appear to be difficult to evaluate in closed form. Perhaps "my given problem" can be solved in other ways. But possibly none of us can do that unless we know what that problem is. – André Nicolas Dec 26 '11 at 19:39
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Ok! Thank you for your assistance! However I should say that these integrals are regard the expansion of function$f(x)=(1+x^{2})^{\frac{1}{4}}sin5x$ in a series of legendre and chebyshev polynomials, so I think that they have one way to solve it. – Roin Dec 27 '11 at 07:48
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Either you should accept the answer if you want or start a bounty (specifically saying what you did not find in these answers) – Jeremy Carlos Mar 15 '12 at 00:00
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$\int_0^1x^n(1+x^2)^\frac{1}{4}\sin5x~dx=\int_0^{\sinh^{-1}1}\sinh^nx(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)~d(\sinh x)=\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)\sinh^nx\cosh^\frac{3}{2}x~dx~,~\int_0^1\dfrac{x^n(1+x^2)^\frac{1}{4}\sin5x}{\sqrt{1-x^2}}dx=\int_0^{\sinh^{-1}1}\dfrac{\sinh^nx(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)}{\sqrt{1-\sinh^2x}}d(\sinh x)=\int_0^{\sinh^{-1}1}\dfrac{\sin(5\sinh x)}{\sqrt{1-\sinh^2x}\sinh^nx\cosh^\frac{3}{2}x}dx$ think about incomplete types of bessel functions – Harry Peter Oct 28 '14 at 16:04
2 Answers
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To refresh the definition here $f(x)$ is an odd function if $f(-x) = -f(x)$, and is even if $f(-x)=f(x)$
Firstly, $$\int_{-1}^1 f(x) \hspace{4pt} dx = 0$$ if $f(x)$ is odd function, and if $f(x)$ is an even function then
$$\int_{-1}^1 f(x) dx = 2 \int_0^1 f(x) dx $$
For the first problem:
$$f(x) = (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x$$
$$ \begin{align*} f(-x) &= (-x^{5} + x^{3} +x)(1+x^{2})^{\frac{1}{4}} \sin5(-x) \\ &= (-x^{5} + x^{3} +x)(1+x^{2})^{\frac{1}{4}} (-\sin 5x)\\ &= (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin 5x = f(x)\\ \end{align*} $$
For the first function if you use the approximation (use high order terms if you want ) $(1+x^2)^{\frac{1}{4}} \approx 1+\frac{1}{4}x^{2}$
The integral is
$$ 2 \int_0^1 (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x dx \approx 2 \int_0^1 (x^{5}-x^{3}-x)(1+\frac{x^2}{4})\sin5x dx$$
which is
$$ 2 \int_0^1 \left( \frac{3}{4} x^5 \sin(5x) -x \sin(5x) - \frac{5x^3}{4} \sin(5x) + \frac{x^7}{4} \sin(5x) \right) dx $$
$$ \approx 2 \left( \frac{98165 \cos(5) -22233 \sin(5)}{312500} \right) = 0.314658 $$
Where this can be verified on Wolfram Alpha
Similarly for the second integral
$$ \int_{-1}^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\sin5x dx = 2 \int_0^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\sin5x dx$$
Using another approximation
$$ \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{\frac{-1}{2}} \approx (1+\frac{x^2}{2}) $$
Therefore
$$ \int_{-1}^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\hspace{4pt}sin5x \hspace{4pt} dx = 2 \int_0^1 (x^{5}-x^{3}-x)(1+\frac{1}{4}x^{2})(1+\frac{1}{2}x^2)\sin5x dx$$
$$ = 2 \int_0^1 \left(\frac{x^9}{8} + \frac{x^7}{4} - \frac{x^5}{8} - \frac{7x^3}{8} -x \right)\sin5x dx $$
$$ = \frac{2}{15625000} \left( (703125x^8 - 481250 x^6 + 186875 x^4 -1730325 x^2 - 486574)\sin5x - 5x(78125 x^8 - 68750x^6 + 37375 x^4-576775x^2 -486574)\cos5x \right) $$
with limits from $0$ to $1$
The value of second integral is therefore $0.406494$
(Note that the simplified function in the second integral is also an even function, and therefore the integral limits are from $0$ to $1$)
Kirthi Raman
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For $\int_{-1}^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$ ,
$\int_{-1}^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$
$=2\int_0^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$
$=2\int_0^1x^5(1+x^2)^\frac{1}{4}\sin5x~dx-2\int_0^1x(1+x^2)^\frac{5}{4}\sin5x~dx$
$=\int_0^1x^4(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)$
$=\int_0^1(1+2x^2+x^4)(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(2+2x^2)(1+x^2)^\frac{1}{4}\sin5x~d(x^2)+\int_0^1(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)$
$=\int_0^1(1+x^2)^\frac{9}{4}\sin5x~d(x^2)-3\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)+\int_0^1(1+x^2)^\frac{1}{4}\sin5x~d(x^2)$
$=\dfrac{4}{13}\int_0^1\sin5x~d\left((1+x^2)^\frac{13}{4}\right)-\dfrac{4}{3}\int_0^1\sin5x~d\left((1+x^2)^\frac{9}{4}\right)+\dfrac{4}{5}\int_0^1\sin5x~d\left((1+x^2)^\frac{5}{4}\right)$
$=\dfrac{4}{13}\left[(1+x^2)^\frac{13}{4}\sin5x\right]_0^1-\dfrac{4}{13}\int_0^1(1+x^2)^\frac{13}{4}~d(\sin5x)-\dfrac{4}{3}\left[(1+x^2)^\frac{9}{4}\sin5x\right]_0^1+\dfrac{4}{3}\int_0^1(1+x^2)^\frac{9}{4}~d(\sin5x)+\dfrac{4}{5}\left[(1+x^2)^\frac{5}{4}\sin5x\right]_0^1-\dfrac{4}{5}\int_0^1(1+x^2)^\frac{5}{4}~d(\sin5x)$
$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^1(1+x^2)^\frac{13}{4}\cos5x~dx+\dfrac{20}{3}\int_0^1(1+x^2)^\frac{9}{4}\cos5x~dx-4\int_0^1(1+x^2)^\frac{5}{4}\cos5x~dx$
$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{13}{4}\cos(5\sinh x)~d(\sinh x)+\dfrac{20}{3}\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{9}{4}\cos(5\sinh x)~d(\sinh x)-4\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{5}{4}\cos(5\sinh x)~d(\sinh x)$
$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{15}{2}x~dx+\dfrac{20}{3}\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{11}{2}x~dx-4\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{7}{2}x~dx$
Which relates to incomplete Bessel function.
For $\int_{-1}^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$ ,
$\int_{-1}^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$
$=2\int_0^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$
$=2\int_0^1\dfrac{x^5(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx-2\int_0^1\dfrac{x(1+x^2)^{\frac{5}{4}}\sin5x}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{x^4(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~d(x^2)-2\int_0^1\dfrac{(1+x^2)^{\frac{5}{4}}\sin5x}{\sqrt{1-x^2}}~d(x^2)$
Harry Peter
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