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It is well known that some definite integrals such as $$\int_{0}^{\pi} \frac{dx}{a+\cos{x}}$$ $$\int_{0}^{\infty} \frac{\sin{x}}{x}dx$$ are solved by using complex analysis techniques. (It uses residue theorem.)

But some of them are proved by only substitution like Harish Chandra Rajpoot answer. $$\int_{0}^{\pi} \log(\sin{x}) dx =-\pi\log{2}$$ (See definite integral without using complex line integral)

Are there any definite integral problems those cannot be soloved without complex line integral tecniques?

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  • No, but there might be easier solutions than using complex line integral techniques. – skyking Sep 02 '15 at 12:01
  • I'm interested, in which sense your first integral converges? – tired Sep 02 '15 at 12:08
  • And yes there are example where i'm quite confident that the only way to get a solution is by complex integration techniques, see here for more information: http://math.stackexchange.com/questions/253910/the-integral-that-stumped-feynman – tired Sep 02 '15 at 12:10
  • @this_is_an_apple The first integral can be done without residues. It can be done with High School Calculus, provided that a>1 because otherwise the integral is divergent (denom has zeros) And that upper limit should not be infinity, perhaps it could be 2pi? – imranfat Sep 02 '15 at 19:07
  • Another very hard nut to crack without some form of complex analysis may be http://math.stackexchange.com/questions/1200613/why-does-int-0-infty-frac-ln-1x-ln2-x-pi2-fracdxx2-give-t – Jack D'Aurizio Sep 02 '15 at 20:44
  • Your first integral is infinite as written. Do you mean the integration interval to be between $[0,2 \pi]$. – Ron Gordon Sep 03 '15 at 00:56

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