How can I evaluate ∫−∞∞cos⁡xcoshxdx and ∫0∞sin⁡xex−1dx

Question

How can I evaluate  ∫−∞∞cos⁡xcoshxdx and ∫0∞sin⁡xex−1dx

Janet Young · Accepted Answer

Expand sech x in an infinite series in e−x:2ex(1+e−2x)=2∑k=0∞(−1)ke−(2k+1)xInterchange the order of integration, and you are left with integrals of the form∫0∞e−axcos⁡xdxwhich can be done using integration by parts, to avoid complex numbers totally, with the result a1+a2. Hence the integral is equal to∑k=0∞(−1)k1+2k1+2k+2k2Okay, and now it is time once again for &#039;&#039;Special Functions and Pray&#039;&#039;. Using the usual tricks, one can express infinite sums in terms of the digamma function as follows: first, write the summand in partial fractions as1+2k1+2k+2k2=12(1k+a+1k+a⋅)where a is the complex root of the denominator, 1+i2. Now we have the identity∑k=0K1k+a=ψ(K+a+1)−ψ(a+1)where ψ=(log⁡Γ)′ as usual. However, this is not good enough: we have to deal with the (−1)k. This we do by splitting into the even and odd cases: some algebra shows the result we want is∑k=0K(−1)kk+a=12(−1)Kψ(a2+K2+1)−12(−1)Kψ(a2+K2+12)−12ψ(a2)+12ψ(a2+12)Now, this sum actually converges by comparison with the alternating harmonic series, the details of which I omit. Hence we can take the limit as k→∞. The problem is what happens to the terms12(−1)K(ψ(a2+K2+1)−ψ(a2+K2+12)),but a slight abuse of Stirling&#039;s approximation shows that this term is O(log⁡n−log⁡(n+12))=o

censoratojk · Accepted Answer

We have  I=∫0∞cos⁡(x)(ex+e−x2)dx=∫0∞2e−x⋅cos⁡(x)1+e−2xdx=∑k=0∞2(−1)k∫0∞e−(2k+1)xcos⁡(x)dx  =2∑k=0∞(−1)kfk  where  fk=∫0∞e−(2k+1)xcos⁡(x)dx=∑i=0∞(−1)l(2l)!∫0∞e−(2k+1)xx2ldx  =∑l=0∞(−1)l(2l)!∫0∞e−tt2ldt(2k+1)2l+1  =(2k+1)(2k+1)2+1  Hence, we have  I=2∑k=0∞(−1)k(2k+1)(2k+1)2+1=π2cosh(π2)

karton · Accepted Answer

You may prove through differential equations that 1cosh⁡x is almost a fixed point of the Fourier transform, meaning that F(1cosh⁡x)=π21cosh⁡πs2 according to Mathematica&#039;s standard choice of normalization constants. Similarly, in the sense of distributions F(cos⁡x)=π2[σ(s−1)+σ(s+1)] hence, by parity, it follows that ∫0+∞cos⁡xcosh⁡xdx=12∫Rcos⁡xcosh⁡xdx=π2⋅1cosh⁡πs2|s=1=π2cosh⁡π2=πeπ/2eπ−1

How can I evaluate \int_{-\infty}^\infty\frac{\cos x}{\cosh x}dx and \int_0^\infty\frac{\sin x}{e^x-1}dx

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New Questions in Calculus 1