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Is there a non-complex number involving method to find the following integral

$$\int_0^\infty\sin(x^n)dx$$

Maybe something using the idea similar to that for evaluating $\frac{\sin x}{x}$ under the same limits. I have no idea about contours and I am hoping that there should be another real method to evaluate it.

YuiTo Cheng
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Sayam Sethi
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  • WA says this here $$\text{ConditionalExpression}\left[\sin \left(\frac{\pi }{2 n}\right) \Gamma \left(1+\frac{1}{n}\right),\Re(n)>1\land \Im(n)=0\right]$$ – Dr. Sonnhard Graubner Feb 18 '19 at 15:10
  • @Dr.SonnhardGraubner that is where my friend found this integral – Sayam Sethi Feb 18 '19 at 15:12
  • I can get this simply with contour integration, but I have tried a couple of real only approaches that have not gotten anywhere. If you want a complex analysis approach, i can supply one. – robjohn Feb 18 '19 at 18:28
  • @robjohn any solution which involves complex numbers is fine as long as it is limited to $cis(\theta)$ no contours please – Sayam Sethi Feb 19 '19 at 02:21
  • @SayamSethi: Since your question is closed, I wrote this answer. It is totally real, no complex analysis (the reflection formula and the Beta function proof cited uses contour integration, but those are well-known and there are real proofs of them). – robjohn Feb 19 '19 at 03:41
  • @robjohn thanks a lot – Sayam Sethi Feb 19 '19 at 03:54
  • @SayamSethi: I have added a non-contour integration proof of the Beta function (and therefore, the reflection formula). – robjohn Feb 19 '19 at 07:34

1 Answers1

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Let $C=\int_0^\infty\cos (x^n) dx,\,S=\int_0^\infty\sin (x^n) dx$ so$$C+iS=\frac{1}{n}\int_0^\infty y^{1/n-1}\exp (iy)dy\\=\frac{1}{n\Gamma(1/n)}\int_0^\infty\int_0^\infty z^{-1/n}\exp [-y(z-i)]dydz\\=\frac{1}{n\Gamma(1/n)}\int_0^\infty\frac{z^{-1/n}(z+i)dz}{z^2+1}.$$You can do the rest with Beta and Gamma functions. Just doing the sine-based integral as requested, the substitution $z=\tan t$ gives$$\frac{1}{n\Gamma(1/n)}\int_0^{\pi/2}\tan^{-1/n} (t)dt=\frac{\pi}{2n\Gamma(1/n)}\sec\left(\frac{\pi}{2n}\right).$$For $n=2$, this reduces to a famous Fresnel integral.

J.G.
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  • Two things are not very clear to me.... What happened to the $y^(1/n-1)$ when you introduced gamma and how did you get the last integral value? – Sayam Sethi Feb 18 '19 at 15:35
  • @SayamSethi For your first question, I used this. The second part is a special case of $2\int_0^{\pi/2}\sin^{2a-1}t\cos^{2b-1}tdt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ with $a+b=1$. – J.G. Feb 18 '19 at 15:54
  • Ok... I missed that the second part is a beta function...however I am not sure I get the first part....I have just learnt the elementary integrals (I am still not in my undergrad), do you mind explaining a little more – Sayam Sethi Feb 18 '19 at 15:58
  • @SayamYethi The definition $\Gamma(s):=\int_0^\infty t^{s-1}\exp -tdt$ implies $\Gamma(s)y^{-s}=\int_0^\infty z^{s-1}\exp -yzdz$ from the substitution $t=yz$. – J.G. Feb 18 '19 at 16:02
  • Lack of parentheses make this hard to parse. – robjohn Feb 18 '19 at 18:37
  • @robjohn Thanks; fixed. – J.G. Feb 18 '19 at 18:55