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Here $A>0$, $w$-real,$\mathtt{i}$-complex. Mathematica gives the answer: $$\frac{1}{2}e^{2\pi w}(\mathtt{i}\pi+2\Gamma(0,2\pi w)-2\Gamma(0,2(1+\mathtt{i}A)\pi w)+2\ln(-\mathtt{i}+A)+2\ln(w)-2\ln(w+iA)) $$ My questions: 1. How to obtain this results without mathematica? 2. Why does this integral grow exponentially as a function of $w$?

Mostly I don't understand why this integral explodes.The real part is:$$\int_{0}^{A}\frac{x\cos(2\pi wx)+\sin(2\pi wx)}{1+x^{2}}dx $$ and geometrically I don't understand why this integral grows so fast

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What do you mean by Gamma[0,2πw]? –  AD. May 5 '11 at 8:26
    
incomplete gamma function (upper) –  Katja May 5 '11 at 8:37
    
@AD and others: For clarification the Mathematica Manual say $\Gamma(a,z_0,z_1)$ is the generalized incomplete gamma function $\Gamma(a,z_0)-\Gamma(a,z_1)$. –  Listing May 5 '11 at 8:37
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For now, I'll only note that your results can be recast in terms of sine and cosine integrals; I'm peering at your problem now, and will see what comes up. –  J. M. May 5 '11 at 10:50
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In particular, have a look at this and this... –  J. M. May 5 '11 at 11:13
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1 Answer

up vote 2 down vote accepted

You have a typo in your post. The first exponential in the answer given by Mathematica should read $e^{-2\pi w}$. In total the solution does NOT grow exponentially. The exponential growth of the $\Gamma$-function is "cured" by the factor $e^{-2\pi w}$.

To show that the integral does not grow exponential, it is possible to obtain an asymptotic expression for $w\to\infty$ without resorting to the explicit expression given in your post. To this end, we use integration by parts $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx = \frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A + \frac{i}{2\pi w} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^2} dx.$$ Repetitive integration by parts yields an asymptotic expansion for $w\to\infty$. The boundary term is the leading term, thus we have $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right).$$

For fun, I also give the next order term (if one is only interested in large $w$ this asymptotic expansion may prove more useful than the exact expression in terms of not so elementary functions). The next integration by parts yields $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx=\frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A - \frac{e^{-2\pi iwx}}{[2\pi w(x-i)]^2} \Biggr|_{x=0}^A- \frac{1}{(2\pi w)^2} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^3} dx,$$ which gives the next term in the asymptotic expansion $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right) - \frac{1}{(2\pi w)^2} \left( 1+ \frac{e^{-2\pi i w A}}{ (A-i)^2}\right). $$

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hm, now I am confused. I have checked,Mathematica gives me $e^{2\pi w}$. I would prefer $e^{-2\pi w}$ too –  Katja May 5 '11 at 11:08
    
@Katja: $e^{2\pi w}$ is wrong. My post shows how the integral should behave for large $w$. –  Fabian May 5 '11 at 11:09
    
@Katja: the two terms which I have given in my post approximate the integral reasonable well for almost all values of $w\gtrsim 1$ and $A$. –  Fabian May 5 '11 at 11:26
    
Tsk, I already upvoted this (and thus cannot upvote it again), but I agree; if $w$ is big, this is a better choice for the numerics. –  J. M. May 5 '11 at 23:47
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