$\displaystyle \int_{0}^{-\infty} \left [ 2^{at} \zeta(bt) - 2^{c-dt} \zeta(t) \right ] \ dt$

If $S = \log_{2}{e} - \log_{4}{e} + \log_{6}{e} - \log_{8}{e} \ldots$ can be expressed as the integral above for positive integers $a,b,c,d$ are positive integers, find the value of $200(a^2 + b^2 + c^2 - \sqrt{8d})$ .

The answer is -200.00000.

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@Kartik Sharma @Rajdeep Dhingra please see my report to verify its validity.

Jake Lai
- 6 years, 1 month ago

Nicely done ! @Kartik Sharma

Rajdeep Dhingra
- 6 years, 1 month ago

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I've been thinking about whether this integral representation is valid or not, considering the Riemann zeta function as the series $\displaystyle \zeta(s) = \sum_{n=1}^{\infty} n^{-s}$ really only applies for $Re(s) > 1$ .

Jake Lai
- 6 years, 1 month ago

It was a nice question to get back into the
**
Brilliant
**
mode ! All this time I have been in the offline mode , doing easy Math Questions in the entrance exams .. :(

You fraud ! :P You copied my method ! Jk , I too solved it similarly but I solved it in 4 minutes , can you beat that ?

Btw , I don't think many people will understand the meaning of
**
lo
**
;)

A Former Brilliant Member
- 6 years, 1 month ago

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Lol! Io, well yeah maybe! But 4 minutes is really good. Welcome back(though partially)!

Kartik Sharma
- 6 years, 1 month ago

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Well , I'm not back :P Rajdeep asked me to solve this question and hence I came ..

I cannot say for sure when I'll be back but I can try to make an appearance after I get into BITS Pilani or MIT .

A Former Brilliant Member
- 6 years, 1 month ago

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@A Former Brilliant Member – Aren't you going to IIT ?

Rajdeep Dhingra
- 6 years, 1 month ago

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@Rajdeep Dhingra – Well , I wanted to go there at first , but I'm getting less marks in Physics in JEE Mains , so I can't get into any of the good IITs ,bombay,madras,kgp or kanpur :(

SO I am studying solely for BITSAT now .

But I K you'll get JEE AIR 1 :D

A Former Brilliant Member
- 6 years, 1 month ago

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@A Former Brilliant Member
–
Thanks,

JEE main Marks don't count in Advance. So if you do good in Advance then you can get a decent IIT

Rajdeep Dhingra
- 6 years, 1 month ago

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@Rajdeep Dhingra – Still , I dk if I'll be able to pull off a decent performance,mostly owing to my JEE Mains Physics scare . But I can always go for BITSAT full on .

Let's see , nothing is in our hands . I'll pray that god stays by my side ^_^

A Former Brilliant Member
- 6 years, 1 month ago

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@A Former Brilliant Member – Best of Luck.

Rajdeep Dhingra
- 6 years, 1 month ago

@A Former Brilliant Member – MIT!?? Massachusetts Institute of Technology?

Kartik Sharma
- 6 years, 1 month ago

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@Kartik Sharma – Nah , Manipal Institute of Tech. , Idk why but my teachers tell it's a good place to go to .

A Former Brilliant Member
- 6 years, 1 month ago

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Well, nice problem!

$f(x) = {log}_{2}e - {log}_{4}e + {log}_{6}e...$

We know that ${log}_{a}b = \frac{ln(b)}{ln(a)}$

Hence, $f(x)$ becomes -

$\displaystyle \sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{ln(2k)}}$

Now, we know that $\displaystyle \int{{a}^{x}} = \frac{{a}^{x}}{ln(a)}$

Io, we have got the natural logarithm in the denominator!

We here would try to guess about the bounds. As we can see, we need $1$ in the numerator and that ${a}^{0} = 1$ . Also, the 2nd bound must make the function zero. Well... what's $ln(0)$ ? You got it right - $-\infty$ , so we have got the other too.

$\displaystyle \frac{1}{ln(2k)} = \int_{-\infty}^{0}{{(2k)}^{x}}$

Well, we can do 2 more things,

$\displaystyle \frac{1}{ln(2k)} = (-1)(-1)\int_{0}^{-\infty}{{(2k)}^{-x}}$

Put in the sum,

$\displaystyle \sum_{k=1}^{\infty}{{(-1)}^{k-1}\int_{0}^{-\infty}{{(2k)}^{-x}}}$

Getting integral sign outside,

$\displaystyle \int_{0}^{-\infty}{\sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{(2k)}^{x}}}}$

$\displaystyle \int_{0}^{-\infty}{{2}^{-x}\sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{k}^{x}}}}$

Now, we will first find out the sum -

We know $\displaystyle \zeta(x) = \sum_{k=1}^{\infty}{\frac{1}{{a}^{x}}}$

Multiplying ${2}^{1-x}$ both sides,

$\displaystyle ({2}^{1-x}\zeta(x) = \sum_{k=1}^{\infty}{\frac{2}{{2a}^{x}}}$

Subtracting these 2 equations(1-2),

$\displaystyle {2}^{-x}({2}^x - 2)\zeta(x) = \sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{a}^{x}}}$

Hence, the function under integral sign equates to

$\displaystyle {2}^{-2x}({2}^{x} - 2)\zeta(x) = {2}^{-x}\zeta(x) - {2}^{1-2x}\zeta(x)$

As a result,

$\displaystyle \int_{0}^{-\infty}{{2}^{-x}\zeta(x) - {2}^{1-2x}\zeta(x)}$